Methods to solve $int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx$












12












$begingroup$


I have a feel this will be a duplicate question. I have had a look around and couldn't find it, so please advise if so.



Here I wish to address the definite integral:



begin{equation}
I = int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx
end{equation}



I have solved it using Feynman's Trick, however I feel it's limited and am hoping to find other methods to solve. Without using Residues, what are some other approaches to this integral?



My method:



begin{equation}
I(t) = int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx
end{equation}



Here $I = I(1)$ and $I(0) = frac{pi}{2}$. Take the derivative under the curve with respect to '$t$' to achieve:



begin{align}
I'(t) &= int_{0}^{infty} frac{-x^2e^{-tx^2}}{x^2 + 1}:dx = -int_{0}^{infty} frac{x^2e^{-tx^2}}{x^2 + 1}:dx \
&= -left[int_{0}^{infty} frac{left(x^2 + 1 - 1right)e^{-tx^2}}{x^2 + 1}:dx right] \
&= -int_{0}^{infty} e^{-tx^2}:dx + int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx \
&= -frac{sqrt{pi}}{2}frac{1}{sqrt{t}} + I(t)
end{align}



And so we arrive at the differential equation:



begin{equation}
I'(t) - I(t) = -frac{sqrt{pi}}{2}frac{1}{sqrt{t}}
end{equation}



Which yields the solution:



begin{equation}
I(t) = frac{pi}{2}e^toperatorname{erfc}left(tright)
end{equation}



Thus,



begin{equation}
I = I(1) int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx = frac{pi}{2}eoperatorname{erfc}(1)
end{equation}



Addendum:



Using the exact method I've employed, you can extend the above integral into a more genealised form:



begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{x^2 + 1}:dx = frac{pi}{2}e^koperatorname{erfc}(sqrt{k})
end{equation}



Addendum 2:
Whilst we are genealising:
begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{ax^2 + b}:dx = frac{pi}{2b}e^Phioperatorname{erfc}(sqrt{Phi})
end{equation}



Where $Phi = frac{kb}{a}$ and $a,b,k in mathbb{R}^{+}$






real-analysis integration definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    something is wrong. $I<int_0^infty e^{-x^2}=sqrtpi/2$. Your answer is greater than this value
    $endgroup$
    – Andrei
    Dec 16 '18 at 2:54










  • $begingroup$
    @Andrei - You are indeed correct, I mistyped. It should be $operatorname{efrc}$ not $operatorname{erf}$. Thank you for the pickup.
    $endgroup$
    – DavidG
    Dec 16 '18 at 2:55
















12












$begingroup$


I have a feel this will be a duplicate question. I have had a look around and couldn't find it, so please advise if so.



Here I wish to address the definite integral:



begin{equation}
I = int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx
end{equation}



I have solved it using Feynman's Trick, however I feel it's limited and am hoping to find other methods to solve. Without using Residues, what are some other approaches to this integral?



My method:



begin{equation}
I(t) = int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx
end{equation}



Here $I = I(1)$ and $I(0) = frac{pi}{2}$. Take the derivative under the curve with respect to '$t$' to achieve:



begin{align}
I'(t) &= int_{0}^{infty} frac{-x^2e^{-tx^2}}{x^2 + 1}:dx = -int_{0}^{infty} frac{x^2e^{-tx^2}}{x^2 + 1}:dx \
&= -left[int_{0}^{infty} frac{left(x^2 + 1 - 1right)e^{-tx^2}}{x^2 + 1}:dx right] \
&= -int_{0}^{infty} e^{-tx^2}:dx + int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx \
&= -frac{sqrt{pi}}{2}frac{1}{sqrt{t}} + I(t)
end{align}



And so we arrive at the differential equation:



begin{equation}
I'(t) - I(t) = -frac{sqrt{pi}}{2}frac{1}{sqrt{t}}
end{equation}



Which yields the solution:



begin{equation}
I(t) = frac{pi}{2}e^toperatorname{erfc}left(tright)
end{equation}



Thus,



begin{equation}
I = I(1) int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx = frac{pi}{2}eoperatorname{erfc}(1)
end{equation}



Addendum:



Using the exact method I've employed, you can extend the above integral into a more genealised form:



begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{x^2 + 1}:dx = frac{pi}{2}e^koperatorname{erfc}(sqrt{k})
end{equation}



Addendum 2:
Whilst we are genealising:
begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{ax^2 + b}:dx = frac{pi}{2b}e^Phioperatorname{erfc}(sqrt{Phi})
end{equation}



Where $Phi = frac{kb}{a}$ and $a,b,k in mathbb{R}^{+}$






real-analysis integration definite-integrals







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    something is wrong. $I<int_0^infty e^{-x^2}=sqrtpi/2$. Your answer is greater than this value
    $endgroup$
    – Andrei
    Dec 16 '18 at 2:54










  • $begingroup$
    @Andrei - You are indeed correct, I mistyped. It should be $operatorname{efrc}$ not $operatorname{erf}$. Thank you for the pickup.
    $endgroup$
    – DavidG
    Dec 16 '18 at 2:55














12












12








12


10



$begingroup$


I have a feel this will be a duplicate question. I have had a look around and couldn't find it, so please advise if so.



Here I wish to address the definite integral:



begin{equation}
I = int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx
end{equation}



I have solved it using Feynman's Trick, however I feel it's limited and am hoping to find other methods to solve. Without using Residues, what are some other approaches to this integral?



My method:



begin{equation}
I(t) = int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx
end{equation}



Here $I = I(1)$ and $I(0) = frac{pi}{2}$. Take the derivative under the curve with respect to '$t$' to achieve:



begin{align}
I'(t) &= int_{0}^{infty} frac{-x^2e^{-tx^2}}{x^2 + 1}:dx = -int_{0}^{infty} frac{x^2e^{-tx^2}}{x^2 + 1}:dx \
&= -left[int_{0}^{infty} frac{left(x^2 + 1 - 1right)e^{-tx^2}}{x^2 + 1}:dx right] \
&= -int_{0}^{infty} e^{-tx^2}:dx + int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx \
&= -frac{sqrt{pi}}{2}frac{1}{sqrt{t}} + I(t)
end{align}



And so we arrive at the differential equation:



begin{equation}
I'(t) - I(t) = -frac{sqrt{pi}}{2}frac{1}{sqrt{t}}
end{equation}



Which yields the solution:



begin{equation}
I(t) = frac{pi}{2}e^toperatorname{erfc}left(tright)
end{equation}



Thus,



begin{equation}
I = I(1) int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx = frac{pi}{2}eoperatorname{erfc}(1)
end{equation}



Addendum:



Using the exact method I've employed, you can extend the above integral into a more genealised form:



begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{x^2 + 1}:dx = frac{pi}{2}e^koperatorname{erfc}(sqrt{k})
end{equation}



Addendum 2:
Whilst we are genealising:
begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{ax^2 + b}:dx = frac{pi}{2b}e^Phioperatorname{erfc}(sqrt{Phi})
end{equation}



Where $Phi = frac{kb}{a}$ and $a,b,k in mathbb{R}^{+}$






real-analysis integration definite-integrals







share|cite|improve this question











$endgroup$




I have a feel this will be a duplicate question. I have had a look around and couldn't find it, so please advise if so.



Here I wish to address the definite integral:



begin{equation}
I = int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx
end{equation}



I have solved it using Feynman's Trick, however I feel it's limited and am hoping to find other methods to solve. Without using Residues, what are some other approaches to this integral?



My method:



begin{equation}
I(t) = int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx
end{equation}



Here $I = I(1)$ and $I(0) = frac{pi}{2}$. Take the derivative under the curve with respect to '$t$' to achieve:



begin{align}
I'(t) &= int_{0}^{infty} frac{-x^2e^{-tx^2}}{x^2 + 1}:dx = -int_{0}^{infty} frac{x^2e^{-tx^2}}{x^2 + 1}:dx \
&= -left[int_{0}^{infty} frac{left(x^2 + 1 - 1right)e^{-tx^2}}{x^2 + 1}:dx right] \
&= -int_{0}^{infty} e^{-tx^2}:dx + int_{0}^{infty} frac{e^{-tx^2}}{x^2 + 1}:dx \
&= -frac{sqrt{pi}}{2}frac{1}{sqrt{t}} + I(t)
end{align}



And so we arrive at the differential equation:



begin{equation}
I'(t) - I(t) = -frac{sqrt{pi}}{2}frac{1}{sqrt{t}}
end{equation}



Which yields the solution:



begin{equation}
I(t) = frac{pi}{2}e^toperatorname{erfc}left(tright)
end{equation}



Thus,



begin{equation}
I = I(1) int_{0}^{infty} frac{e^{-x^2}}{x^2 + 1}:dx = frac{pi}{2}eoperatorname{erfc}(1)
end{equation}



Addendum:



Using the exact method I've employed, you can extend the above integral into a more genealised form:



begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{x^2 + 1}:dx = frac{pi}{2}e^koperatorname{erfc}(sqrt{k})
end{equation}



Addendum 2:
Whilst we are genealising:
begin{equation}
I = int_{0}^{infty} frac{e^{-kx^2}}{ax^2 + b}:dx = frac{pi}{2b}e^Phioperatorname{erfc}(sqrt{Phi})
end{equation}



Where $Phi = frac{kb}{a}$ and $a,b,k in mathbb{R}^{+}$






real-analysis integration definite-integrals




real-analysis integration definite-integrals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 16 '18 at 3:38







DavidG

















asked Dec 16 '18 at 2:42









DavidGDavidG

2,1121723




2,1121723








  • 1




    $begingroup$
    something is wrong. $I<int_0^infty e^{-x^2}=sqrtpi/2$. Your answer is greater than this value
    $endgroup$
    – Andrei
    Dec 16 '18 at 2:54










  • $begingroup$
    @Andrei - You are indeed correct, I mistyped. It should be $operatorname{efrc}$ not $operatorname{erf}$. Thank you for the pickup.
    $endgroup$
    – DavidG
    Dec 16 '18 at 2:55














  • 1




    $begingroup$
    something is wrong. $I<int_0^infty e^{-x^2}=sqrtpi/2$. Your answer is greater than this value
    $endgroup$
    – Andrei
    Dec 16 '18 at 2:54










  • $begingroup$
    @Andrei - You are indeed correct, I mistyped. It should be $operatorname{efrc}$ not $operatorname{erf}$. Thank you for the pickup.
    $endgroup$
    – DavidG
    Dec 16 '18 at 2:55








1




1




$begingroup$
something is wrong. $I<int_0^infty e^{-x^2}=sqrtpi/2$. Your answer is greater than this value
$endgroup$
– Andrei
Dec 16 '18 at 2:54




$begingroup$
something is wrong. $I<int_0^infty e^{-x^2}=sqrtpi/2$. Your answer is greater than this value
$endgroup$
– Andrei
Dec 16 '18 at 2:54












$begingroup$
@Andrei - You are indeed correct, I mistyped. It should be $operatorname{efrc}$ not $operatorname{erf}$. Thank you for the pickup.
$endgroup$
– DavidG
Dec 16 '18 at 2:55




$begingroup$
@Andrei - You are indeed correct, I mistyped. It should be $operatorname{efrc}$ not $operatorname{erf}$. Thank you for the pickup.
$endgroup$
– DavidG
Dec 16 '18 at 2:55










2 Answers
2






active

oldest

votes


















17












$begingroup$

You can use Plancherel's theorem. Note that
$$
2I = int_{-infty}^{infty} frac{e^{-x^2}}{x^2 + 1}dx.
$$Let $f(x) = e^{-x^2}$ and $g(x) = frac{1}{1+x^2}$. Then we have
$$
widehat{f}(xi) = sqrt{pi}e^{-pi^2xi^2},
$$ and
$$
widehat{g}(xi) = pi e^{-2pi|xi|}.
$$ By Plancherel's theorem, we have
$$begin{eqnarray}
int_{-infty}^{infty} f(x)g(x)dx&=&int_{-infty}^{infty} widehat{f}(xi)widehat{g}(xi)dxi\&=&pi^{frac{3}{2}}int_{-infty}^{infty}e^{-pi^2xi^2-2pi|xi|}dxi\
&=&2pi^{frac{3}{2}}int_{0}^{infty}e^{-pi^2xi^2-2pixi}dxi\
&=&2pi^{frac{3}{2}}eint_{frac{1}{pi}}^{infty}e^{-pi^2xi^2}dxi\
&=&2pi^{frac{1}{2}}eint_{1}^{infty}e^{-xi^2}dxi = pi e operatorname{erfc}(1).
end{eqnarray}$$
This gives $I = frac{pi}{2}e operatorname{erfc}(1).$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @Song! I was not aware of Plancherel's theorem :-)
    $endgroup$
    – DavidG
    Dec 16 '18 at 3:17










  • $begingroup$
    With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:38






  • 1




    $begingroup$
    Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
    $endgroup$
    – Song
    Dec 20 '18 at 2:20





















7












$begingroup$

Here is a method that employs the old trick of converting the integral into a double integral.



Observe that
$$frac{1}{1 + x^2} = int_0^infty e^{-u(1 + x^2)} , du.$$
So your integral can be rewritten as
$$I = int_0^infty e^{-x^2} int_0^infty e^{-u(1 + x^2)} , du , dx.$$
or
$$I = int_0^infty e^{-u} int_0^infty e^{-(1 + u)x^2} , dx , du,$$
on changing the order of integration.



Enforcing a substitution of $x mapsto x/sqrt{1 + u}$ gives
$$I = int_0^infty frac{e^{-u}}{sqrt{1 + u}} int_0^infty e^{-x^2} , dx = frac{sqrt{pi}}{2} int_0^infty frac{e^{-u}}{sqrt{1 + u}} , du.$$



Next exforcing a substitution of $u mapsto u^2 - 1$ gives
$$I = sqrt{pi} e int_1^infty e^{-u^2} , du = sqrt{pi} e cdot frac{sqrt{pi}}{2} text{erf} (1) = frac{pi e}{2} text{erf} (1),$$
as expected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Love it. Still developing my skills with respect to integral ‘expansions’.
    $endgroup$
    – DavidG
    Dec 16 '18 at 10:00










  • $begingroup$
    Is another one of those good techniques to have handy in the integration toolbox.
    $endgroup$
    – omegadot
    Dec 16 '18 at 11:11










  • $begingroup$
    Absolutely. Great technique.
    $endgroup$
    – DavidG
    Dec 16 '18 at 11:12











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









17












$begingroup$

You can use Plancherel's theorem. Note that
$$
2I = int_{-infty}^{infty} frac{e^{-x^2}}{x^2 + 1}dx.
$$Let $f(x) = e^{-x^2}$ and $g(x) = frac{1}{1+x^2}$. Then we have
$$
widehat{f}(xi) = sqrt{pi}e^{-pi^2xi^2},
$$ and
$$
widehat{g}(xi) = pi e^{-2pi|xi|}.
$$ By Plancherel's theorem, we have
$$begin{eqnarray}
int_{-infty}^{infty} f(x)g(x)dx&=&int_{-infty}^{infty} widehat{f}(xi)widehat{g}(xi)dxi\&=&pi^{frac{3}{2}}int_{-infty}^{infty}e^{-pi^2xi^2-2pi|xi|}dxi\
&=&2pi^{frac{3}{2}}int_{0}^{infty}e^{-pi^2xi^2-2pixi}dxi\
&=&2pi^{frac{3}{2}}eint_{frac{1}{pi}}^{infty}e^{-pi^2xi^2}dxi\
&=&2pi^{frac{1}{2}}eint_{1}^{infty}e^{-xi^2}dxi = pi e operatorname{erfc}(1).
end{eqnarray}$$
This gives $I = frac{pi}{2}e operatorname{erfc}(1).$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @Song! I was not aware of Plancherel's theorem :-)
    $endgroup$
    – DavidG
    Dec 16 '18 at 3:17










  • $begingroup$
    With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:38






  • 1




    $begingroup$
    Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
    $endgroup$
    – Song
    Dec 20 '18 at 2:20


















17












$begingroup$

You can use Plancherel's theorem. Note that
$$
2I = int_{-infty}^{infty} frac{e^{-x^2}}{x^2 + 1}dx.
$$Let $f(x) = e^{-x^2}$ and $g(x) = frac{1}{1+x^2}$. Then we have
$$
widehat{f}(xi) = sqrt{pi}e^{-pi^2xi^2},
$$ and
$$
widehat{g}(xi) = pi e^{-2pi|xi|}.
$$ By Plancherel's theorem, we have
$$begin{eqnarray}
int_{-infty}^{infty} f(x)g(x)dx&=&int_{-infty}^{infty} widehat{f}(xi)widehat{g}(xi)dxi\&=&pi^{frac{3}{2}}int_{-infty}^{infty}e^{-pi^2xi^2-2pi|xi|}dxi\
&=&2pi^{frac{3}{2}}int_{0}^{infty}e^{-pi^2xi^2-2pixi}dxi\
&=&2pi^{frac{3}{2}}eint_{frac{1}{pi}}^{infty}e^{-pi^2xi^2}dxi\
&=&2pi^{frac{1}{2}}eint_{1}^{infty}e^{-xi^2}dxi = pi e operatorname{erfc}(1).
end{eqnarray}$$
This gives $I = frac{pi}{2}e operatorname{erfc}(1).$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thanks @Song! I was not aware of Plancherel's theorem :-)
    $endgroup$
    – DavidG
    Dec 16 '18 at 3:17










  • $begingroup$
    With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:38






  • 1




    $begingroup$
    Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
    $endgroup$
    – Song
    Dec 20 '18 at 2:20
















17












17








17





$begingroup$

You can use Plancherel's theorem. Note that
$$
2I = int_{-infty}^{infty} frac{e^{-x^2}}{x^2 + 1}dx.
$$Let $f(x) = e^{-x^2}$ and $g(x) = frac{1}{1+x^2}$. Then we have
$$
widehat{f}(xi) = sqrt{pi}e^{-pi^2xi^2},
$$ and
$$
widehat{g}(xi) = pi e^{-2pi|xi|}.
$$ By Plancherel's theorem, we have
$$begin{eqnarray}
int_{-infty}^{infty} f(x)g(x)dx&=&int_{-infty}^{infty} widehat{f}(xi)widehat{g}(xi)dxi\&=&pi^{frac{3}{2}}int_{-infty}^{infty}e^{-pi^2xi^2-2pi|xi|}dxi\
&=&2pi^{frac{3}{2}}int_{0}^{infty}e^{-pi^2xi^2-2pixi}dxi\
&=&2pi^{frac{3}{2}}eint_{frac{1}{pi}}^{infty}e^{-pi^2xi^2}dxi\
&=&2pi^{frac{1}{2}}eint_{1}^{infty}e^{-xi^2}dxi = pi e operatorname{erfc}(1).
end{eqnarray}$$
This gives $I = frac{pi}{2}e operatorname{erfc}(1).$






share|cite|improve this answer











$endgroup$



You can use Plancherel's theorem. Note that
$$
2I = int_{-infty}^{infty} frac{e^{-x^2}}{x^2 + 1}dx.
$$Let $f(x) = e^{-x^2}$ and $g(x) = frac{1}{1+x^2}$. Then we have
$$
widehat{f}(xi) = sqrt{pi}e^{-pi^2xi^2},
$$ and
$$
widehat{g}(xi) = pi e^{-2pi|xi|}.
$$ By Plancherel's theorem, we have
$$begin{eqnarray}
int_{-infty}^{infty} f(x)g(x)dx&=&int_{-infty}^{infty} widehat{f}(xi)widehat{g}(xi)dxi\&=&pi^{frac{3}{2}}int_{-infty}^{infty}e^{-pi^2xi^2-2pi|xi|}dxi\
&=&2pi^{frac{3}{2}}int_{0}^{infty}e^{-pi^2xi^2-2pixi}dxi\
&=&2pi^{frac{3}{2}}eint_{frac{1}{pi}}^{infty}e^{-pi^2xi^2}dxi\
&=&2pi^{frac{1}{2}}eint_{1}^{infty}e^{-xi^2}dxi = pi e operatorname{erfc}(1).
end{eqnarray}$$
This gives $I = frac{pi}{2}e operatorname{erfc}(1).$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 16 '18 at 3:21

























answered Dec 16 '18 at 3:15









SongSong

13.2k632




13.2k632












  • $begingroup$
    Thanks @Song! I was not aware of Plancherel's theorem :-)
    $endgroup$
    – DavidG
    Dec 16 '18 at 3:17










  • $begingroup$
    With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:38






  • 1




    $begingroup$
    Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
    $endgroup$
    – Song
    Dec 20 '18 at 2:20




















  • $begingroup$
    Thanks @Song! I was not aware of Plancherel's theorem :-)
    $endgroup$
    – DavidG
    Dec 16 '18 at 3:17










  • $begingroup$
    With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
    $endgroup$
    – DavidG
    Dec 20 '18 at 0:38






  • 1




    $begingroup$
    Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
    $endgroup$
    – Song
    Dec 20 '18 at 2:20


















$begingroup$
Thanks @Song! I was not aware of Plancherel's theorem :-)
$endgroup$
– DavidG
Dec 16 '18 at 3:17




$begingroup$
Thanks @Song! I was not aware of Plancherel's theorem :-)
$endgroup$
– DavidG
Dec 16 '18 at 3:17












$begingroup$
With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
$endgroup$
– DavidG
Dec 20 '18 at 0:38




$begingroup$
With respect (as I loved the solution), but if I apply Plancherel's theorem - am I not using residues in disguise? (again, love the solution).
$endgroup$
– DavidG
Dec 20 '18 at 0:38




1




1




$begingroup$
Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
$endgroup$
– Song
Dec 20 '18 at 2:20






$begingroup$
Your concern is legitimate. Perhaps the quickest way to evaluate $widehat{f}$ and $widehat{g}$ is to use complex analysis tools. But the method is still real analytic. We can evaluate $int_{-infty}^infty e^{zx}e^{-x^2}dx,;zinmathbb{C}$ easily using power series expansion. About $widehat{g}$, note that Fourier transform admits the inversion formula. So what we should check is only whether $int_{-infty}^inftywidehat{g}(xi)e^{2 pi ixi x}dxi=g(x)$, and it can be checked directly by evaluating the integral separately on $(-infty,0)$ and $(0,infty)$.
$endgroup$
– Song
Dec 20 '18 at 2:20













7












$begingroup$

Here is a method that employs the old trick of converting the integral into a double integral.



Observe that
$$frac{1}{1 + x^2} = int_0^infty e^{-u(1 + x^2)} , du.$$
So your integral can be rewritten as
$$I = int_0^infty e^{-x^2} int_0^infty e^{-u(1 + x^2)} , du , dx.$$
or
$$I = int_0^infty e^{-u} int_0^infty e^{-(1 + u)x^2} , dx , du,$$
on changing the order of integration.



Enforcing a substitution of $x mapsto x/sqrt{1 + u}$ gives
$$I = int_0^infty frac{e^{-u}}{sqrt{1 + u}} int_0^infty e^{-x^2} , dx = frac{sqrt{pi}}{2} int_0^infty frac{e^{-u}}{sqrt{1 + u}} , du.$$



Next exforcing a substitution of $u mapsto u^2 - 1$ gives
$$I = sqrt{pi} e int_1^infty e^{-u^2} , du = sqrt{pi} e cdot frac{sqrt{pi}}{2} text{erf} (1) = frac{pi e}{2} text{erf} (1),$$
as expected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Love it. Still developing my skills with respect to integral ‘expansions’.
    $endgroup$
    – DavidG
    Dec 16 '18 at 10:00










  • $begingroup$
    Is another one of those good techniques to have handy in the integration toolbox.
    $endgroup$
    – omegadot
    Dec 16 '18 at 11:11










  • $begingroup$
    Absolutely. Great technique.
    $endgroup$
    – DavidG
    Dec 16 '18 at 11:12
















7












$begingroup$

Here is a method that employs the old trick of converting the integral into a double integral.



Observe that
$$frac{1}{1 + x^2} = int_0^infty e^{-u(1 + x^2)} , du.$$
So your integral can be rewritten as
$$I = int_0^infty e^{-x^2} int_0^infty e^{-u(1 + x^2)} , du , dx.$$
or
$$I = int_0^infty e^{-u} int_0^infty e^{-(1 + u)x^2} , dx , du,$$
on changing the order of integration.



Enforcing a substitution of $x mapsto x/sqrt{1 + u}$ gives
$$I = int_0^infty frac{e^{-u}}{sqrt{1 + u}} int_0^infty e^{-x^2} , dx = frac{sqrt{pi}}{2} int_0^infty frac{e^{-u}}{sqrt{1 + u}} , du.$$



Next exforcing a substitution of $u mapsto u^2 - 1$ gives
$$I = sqrt{pi} e int_1^infty e^{-u^2} , du = sqrt{pi} e cdot frac{sqrt{pi}}{2} text{erf} (1) = frac{pi e}{2} text{erf} (1),$$
as expected.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Love it. Still developing my skills with respect to integral ‘expansions’.
    $endgroup$
    – DavidG
    Dec 16 '18 at 10:00










  • $begingroup$
    Is another one of those good techniques to have handy in the integration toolbox.
    $endgroup$
    – omegadot
    Dec 16 '18 at 11:11










  • $begingroup$
    Absolutely. Great technique.
    $endgroup$
    – DavidG
    Dec 16 '18 at 11:12














7












7








7





$begingroup$

Here is a method that employs the old trick of converting the integral into a double integral.



Observe that
$$frac{1}{1 + x^2} = int_0^infty e^{-u(1 + x^2)} , du.$$
So your integral can be rewritten as
$$I = int_0^infty e^{-x^2} int_0^infty e^{-u(1 + x^2)} , du , dx.$$
or
$$I = int_0^infty e^{-u} int_0^infty e^{-(1 + u)x^2} , dx , du,$$
on changing the order of integration.



Enforcing a substitution of $x mapsto x/sqrt{1 + u}$ gives
$$I = int_0^infty frac{e^{-u}}{sqrt{1 + u}} int_0^infty e^{-x^2} , dx = frac{sqrt{pi}}{2} int_0^infty frac{e^{-u}}{sqrt{1 + u}} , du.$$



Next exforcing a substitution of $u mapsto u^2 - 1$ gives
$$I = sqrt{pi} e int_1^infty e^{-u^2} , du = sqrt{pi} e cdot frac{sqrt{pi}}{2} text{erf} (1) = frac{pi e}{2} text{erf} (1),$$
as expected.






share|cite|improve this answer









$endgroup$



Here is a method that employs the old trick of converting the integral into a double integral.



Observe that
$$frac{1}{1 + x^2} = int_0^infty e^{-u(1 + x^2)} , du.$$
So your integral can be rewritten as
$$I = int_0^infty e^{-x^2} int_0^infty e^{-u(1 + x^2)} , du , dx.$$
or
$$I = int_0^infty e^{-u} int_0^infty e^{-(1 + u)x^2} , dx , du,$$
on changing the order of integration.



Enforcing a substitution of $x mapsto x/sqrt{1 + u}$ gives
$$I = int_0^infty frac{e^{-u}}{sqrt{1 + u}} int_0^infty e^{-x^2} , dx = frac{sqrt{pi}}{2} int_0^infty frac{e^{-u}}{sqrt{1 + u}} , du.$$



Next exforcing a substitution of $u mapsto u^2 - 1$ gives
$$I = sqrt{pi} e int_1^infty e^{-u^2} , du = sqrt{pi} e cdot frac{sqrt{pi}}{2} text{erf} (1) = frac{pi e}{2} text{erf} (1),$$
as expected.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 16 '18 at 9:56









omegadotomegadot

5,9222828




5,9222828












  • $begingroup$
    Love it. Still developing my skills with respect to integral ‘expansions’.
    $endgroup$
    – DavidG
    Dec 16 '18 at 10:00










  • $begingroup$
    Is another one of those good techniques to have handy in the integration toolbox.
    $endgroup$
    – omegadot
    Dec 16 '18 at 11:11










  • $begingroup$
    Absolutely. Great technique.
    $endgroup$
    – DavidG
    Dec 16 '18 at 11:12


















  • $begingroup$
    Love it. Still developing my skills with respect to integral ‘expansions’.
    $endgroup$
    – DavidG
    Dec 16 '18 at 10:00










  • $begingroup$
    Is another one of those good techniques to have handy in the integration toolbox.
    $endgroup$
    – omegadot
    Dec 16 '18 at 11:11










  • $begingroup$
    Absolutely. Great technique.
    $endgroup$
    – DavidG
    Dec 16 '18 at 11:12
















$begingroup$
Love it. Still developing my skills with respect to integral ‘expansions’.
$endgroup$
– DavidG
Dec 16 '18 at 10:00




$begingroup$
Love it. Still developing my skills with respect to integral ‘expansions’.
$endgroup$
– DavidG
Dec 16 '18 at 10:00












$begingroup$
Is another one of those good techniques to have handy in the integration toolbox.
$endgroup$
– omegadot
Dec 16 '18 at 11:11




$begingroup$
Is another one of those good techniques to have handy in the integration toolbox.
$endgroup$
– omegadot
Dec 16 '18 at 11:11












$begingroup$
Absolutely. Great technique.
$endgroup$
– DavidG
Dec 16 '18 at 11:12




$begingroup$
Absolutely. Great technique.
$endgroup$
– DavidG
Dec 16 '18 at 11:12


















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