Ytukyg

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 $a,b,k in mathbb{R}^{+}$

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 $I = frac{pi}{2}e operatorname{erfc}(1).$

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.