Finding conditions to an equality in an integral












0












$begingroup$



First of all, a happy new year to everybody.




Well, I'm working on a problem. That problem stated that I've to find the conditions for which this equality holds:



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}x=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag1$$




My work:



I proved that it is generally true for positive real numbers (atleast I think I did):



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}xtag2$$



Using the 'evaluating integrals over the positive real axis' of the Laplace transform, we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftymathcal{L}_xleft[sinleft(epsiloncdot xright)right]_{left(text{s}right)}cdotmathcal{L}_x^{-1}left[frac{1}{x^text{n}}right]_{left(text{s}right)}spacetext{d}text{s}tag3$$



Using the 'table of selected Laplace transforms', we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftyfrac{epsilon}{epsilon^2+text{s}^2}cdotfrac{text{s}^{text{n}-1}}{Gammaleft(text{n}right)}spacetext{d}text{s}=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}tag4$$



Using this answer of mine we can write:



$$mathcal{I}_text{n}left(epsilonright)=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag5$$






integration trigonometry definite-integrals laplace-transform gamma-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you take $epsilon = 1$ and $n = 3$ this integral diverges, but the rhs is finite. I think there must be a restriction on $n$
    $endgroup$
    – caverac
    Jan 2 at 0:23












  • $begingroup$
    @caverac I would like to know what that restriction is ;)
    $endgroup$
    – Jan
    Jan 2 at 0:32






  • 1




    $begingroup$
    Sorry, what I mean is: it is definitely not true for all positive real numbers. I just ran some numerical tests, and it seems that the integral does not converge for $n geq 2$. Not a proof, but it may give you something to start (?)
    $endgroup$
    – caverac
    Jan 2 at 0:37










  • $begingroup$
    The integral converges for $epsilon in mathbb R land 0 < operatorname{Re} n < 2$, otherwise there will be divergence either when the lower limit of integration approaches $0$ or when the upper limit approaches infinity. The equality holds under the additional condition $epsilon > 0$.
    $endgroup$
    – Maxim
    Jan 2 at 3:21
















0












$begingroup$



First of all, a happy new year to everybody.




Well, I'm working on a problem. That problem stated that I've to find the conditions for which this equality holds:



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}x=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag1$$




My work:



I proved that it is generally true for positive real numbers (atleast I think I did):



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}xtag2$$



Using the 'evaluating integrals over the positive real axis' of the Laplace transform, we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftymathcal{L}_xleft[sinleft(epsiloncdot xright)right]_{left(text{s}right)}cdotmathcal{L}_x^{-1}left[frac{1}{x^text{n}}right]_{left(text{s}right)}spacetext{d}text{s}tag3$$



Using the 'table of selected Laplace transforms', we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftyfrac{epsilon}{epsilon^2+text{s}^2}cdotfrac{text{s}^{text{n}-1}}{Gammaleft(text{n}right)}spacetext{d}text{s}=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}tag4$$



Using this answer of mine we can write:



$$mathcal{I}_text{n}left(epsilonright)=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag5$$






integration trigonometry definite-integrals laplace-transform gamma-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    If you take $epsilon = 1$ and $n = 3$ this integral diverges, but the rhs is finite. I think there must be a restriction on $n$
    $endgroup$
    – caverac
    Jan 2 at 0:23












  • $begingroup$
    @caverac I would like to know what that restriction is ;)
    $endgroup$
    – Jan
    Jan 2 at 0:32






  • 1




    $begingroup$
    Sorry, what I mean is: it is definitely not true for all positive real numbers. I just ran some numerical tests, and it seems that the integral does not converge for $n geq 2$. Not a proof, but it may give you something to start (?)
    $endgroup$
    – caverac
    Jan 2 at 0:37










  • $begingroup$
    The integral converges for $epsilon in mathbb R land 0 < operatorname{Re} n < 2$, otherwise there will be divergence either when the lower limit of integration approaches $0$ or when the upper limit approaches infinity. The equality holds under the additional condition $epsilon > 0$.
    $endgroup$
    – Maxim
    Jan 2 at 3:21














0












0








0


2



$begingroup$



First of all, a happy new year to everybody.




Well, I'm working on a problem. That problem stated that I've to find the conditions for which this equality holds:



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}x=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag1$$




My work:



I proved that it is generally true for positive real numbers (atleast I think I did):



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}xtag2$$



Using the 'evaluating integrals over the positive real axis' of the Laplace transform, we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftymathcal{L}_xleft[sinleft(epsiloncdot xright)right]_{left(text{s}right)}cdotmathcal{L}_x^{-1}left[frac{1}{x^text{n}}right]_{left(text{s}right)}spacetext{d}text{s}tag3$$



Using the 'table of selected Laplace transforms', we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftyfrac{epsilon}{epsilon^2+text{s}^2}cdotfrac{text{s}^{text{n}-1}}{Gammaleft(text{n}right)}spacetext{d}text{s}=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}tag4$$



Using this answer of mine we can write:



$$mathcal{I}_text{n}left(epsilonright)=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag5$$






integration trigonometry definite-integrals laplace-transform gamma-function







share|cite|improve this question









$endgroup$





First of all, a happy new year to everybody.




Well, I'm working on a problem. That problem stated that I've to find the conditions for which this equality holds:



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}x=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag1$$




My work:



I proved that it is generally true for positive real numbers (atleast I think I did):



$$mathcal{I}_text{n}left(epsilonright):=int_0^inftyfrac{sinleft(epsiloncdot xright)}{x^text{n}}spacetext{d}xtag2$$



Using the 'evaluating integrals over the positive real axis' of the Laplace transform, we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftymathcal{L}_xleft[sinleft(epsiloncdot xright)right]_{left(text{s}right)}cdotmathcal{L}_x^{-1}left[frac{1}{x^text{n}}right]_{left(text{s}right)}spacetext{d}text{s}tag3$$



Using the 'table of selected Laplace transforms', we can write:



$$mathcal{I}_text{n}left(epsilonright)=int_0^inftyfrac{epsilon}{epsilon^2+text{s}^2}cdotfrac{text{s}^{text{n}-1}}{Gammaleft(text{n}right)}spacetext{d}text{s}=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}tag4$$



Using this answer of mine we can write:



$$mathcal{I}_text{n}left(epsilonright)=frac{epsilon}{Gammaleft(text{n}right)}cdotint_0^inftyfrac{text{s}^{text{n}-1}}{epsilon^2+text{s}^2}spacetext{d}text{s}=frac{pi}{2}cdotfrac{epsilon^{text{n}-1}}{Gammaleft(text{n}right)}cdotcscleft(text{n}cdotfrac{pi}{2}right)tag5$$






integration trigonometry definite-integrals laplace-transform gamma-function




integration trigonometry definite-integrals laplace-transform gamma-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 1 at 21:11









JanJan

22.1k31440




22.1k31440












  • $begingroup$
    If you take $epsilon = 1$ and $n = 3$ this integral diverges, but the rhs is finite. I think there must be a restriction on $n$
    $endgroup$
    – caverac
    Jan 2 at 0:23












  • $begingroup$
    @caverac I would like to know what that restriction is ;)
    $endgroup$
    – Jan
    Jan 2 at 0:32






  • 1




    $begingroup$
    Sorry, what I mean is: it is definitely not true for all positive real numbers. I just ran some numerical tests, and it seems that the integral does not converge for $n geq 2$. Not a proof, but it may give you something to start (?)
    $endgroup$
    – caverac
    Jan 2 at 0:37










  • $begingroup$
    The integral converges for $epsilon in mathbb R land 0 < operatorname{Re} n < 2$, otherwise there will be divergence either when the lower limit of integration approaches $0$ or when the upper limit approaches infinity. The equality holds under the additional condition $epsilon > 0$.
    $endgroup$
    – Maxim
    Jan 2 at 3:21


















  • $begingroup$
    If you take $epsilon = 1$ and $n = 3$ this integral diverges, but the rhs is finite. I think there must be a restriction on $n$
    $endgroup$
    – caverac
    Jan 2 at 0:23












  • $begingroup$
    @caverac I would like to know what that restriction is ;)
    $endgroup$
    – Jan
    Jan 2 at 0:32






  • 1




    $begingroup$
    Sorry, what I mean is: it is definitely not true for all positive real numbers. I just ran some numerical tests, and it seems that the integral does not converge for $n geq 2$. Not a proof, but it may give you something to start (?)
    $endgroup$
    – caverac
    Jan 2 at 0:37










  • $begingroup$
    The integral converges for $epsilon in mathbb R land 0 < operatorname{Re} n < 2$, otherwise there will be divergence either when the lower limit of integration approaches $0$ or when the upper limit approaches infinity. The equality holds under the additional condition $epsilon > 0$.
    $endgroup$
    – Maxim
    Jan 2 at 3:21
















$begingroup$
If you take $epsilon = 1$ and $n = 3$ this integral diverges, but the rhs is finite. I think there must be a restriction on $n$
$endgroup$
– caverac
Jan 2 at 0:23






$begingroup$
If you take $epsilon = 1$ and $n = 3$ this integral diverges, but the rhs is finite. I think there must be a restriction on $n$
$endgroup$
– caverac
Jan 2 at 0:23














$begingroup$
@caverac I would like to know what that restriction is ;)
$endgroup$
– Jan
Jan 2 at 0:32




$begingroup$
@caverac I would like to know what that restriction is ;)
$endgroup$
– Jan
Jan 2 at 0:32




1




1




$begingroup$
Sorry, what I mean is: it is definitely not true for all positive real numbers. I just ran some numerical tests, and it seems that the integral does not converge for $n geq 2$. Not a proof, but it may give you something to start (?)
$endgroup$
– caverac
Jan 2 at 0:37




$begingroup$
Sorry, what I mean is: it is definitely not true for all positive real numbers. I just ran some numerical tests, and it seems that the integral does not converge for $n geq 2$. Not a proof, but it may give you something to start (?)
$endgroup$
– caverac
Jan 2 at 0:37












$begingroup$
The integral converges for $epsilon in mathbb R land 0 < operatorname{Re} n < 2$, otherwise there will be divergence either when the lower limit of integration approaches $0$ or when the upper limit approaches infinity. The equality holds under the additional condition $epsilon > 0$.
$endgroup$
– Maxim
Jan 2 at 3:21




$begingroup$
The integral converges for $epsilon in mathbb R land 0 < operatorname{Re} n < 2$, otherwise there will be divergence either when the lower limit of integration approaches $0$ or when the upper limit approaches infinity. The equality holds under the additional condition $epsilon > 0$.
$endgroup$
– Maxim
Jan 2 at 3:21










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