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As the title says, is there a closed form for $$ \int_0^\infty \frac{\sin(x)}{x^2+a^2} dx \,? $$

The one with $\cos(x)$ instead of $\sin(x)$ can be calculated via a simple application of the Residue theorem, but the computation uses the fact that the function is even, so the same trick does not work for this one.

Clarification In case this was unclear, I do not care about the method used to calculate this integral.

N. S.
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  • Hint: $$\sin(x) = \operatorname{Im}e^{ix}.$$ – vitamin d Feb 23 '21 at 18:13
  • @vitamind I do not see how that does any difference. I think that would help if the integral would be the PV over the reals, but it is not. – N. S. Feb 23 '21 at 18:14
  • I can not see why there is a problem with the the integration domain. The countour path has to be a little different. – vitamin d Feb 23 '21 at 18:18
  • @vitamind That is the problem, I cannot find a good contour path. Note that since $e^0=1$, switching to the exponential seems to create issues with the extra piece of the contour path. – N. S. Feb 23 '21 at 18:20
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    @vitamind For $\cos(x)/(x^2+a^2)$ we can compute the integral on the whole of real line by the residue at $ai$, and halve it to give 0 to infinity. This trick doesn't work here for $\sin$ instead. – user10354138 Feb 23 '21 at 18:22
  • I think if you choose a clever path it might work. Maybe you are right and it is not possible. – vitamin d Feb 23 '21 at 18:24
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    Use, on the result here the Feynman's trick (differentiation under the integral sign with respect to parameter $c$) – Jean Marie Feb 23 '21 at 18:30
  • Can we use identity $\cos x=\sin(\frac{\pi}{2}-x)$ and the fact that we can calculate integral with the cosine? – Vasili Feb 23 '21 at 18:31
  • I'm not sure about this path: Starting at $0$ going up to $ai-i\varepsilon$ semi-cirle that includes $ai$ to $ai+i\varepsilon$ going up to $iR$ and coming back to the real axis with a quarter circle with radius $R$ and then back to the origin. – vitamin d Feb 23 '21 at 18:39
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    @vitamind That doesn't work -- along the imaginary axis the denominator becomes $a^2-x^2$ instead of $x^2+a^2$ and there is no easy way to relate them. – user10354138 Feb 23 '21 at 18:42
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    @user10354138 I think an undesired factor of $x$ comes up if we try your $\int_0^1$ suggestion. – J.G. Feb 23 '21 at 18:52
  • There might be a way to derive an ODE for this integral, like in the following answer https://math.stackexchange.com/a/1841104/269624 – Yuriy S Feb 23 '21 at 18:54
  • @YuriyS I think he's trying to find a way via complex analysis – vitamin d Feb 23 '21 at 18:55
  • By the way, WA gives an ugly closed form in terms of exponential integrals, even for a=1, so the answer is yes, there's a closed form, but not elementary – Yuriy S Feb 23 '21 at 18:57
  • Perhaps you could use $$\int_0^\infty \frac{\sin xt}{t^2+a^2}dt=\mathcal{L}\left{\frac{\log(a/s)}{a^2-s^2}\right}(x),$$ where $\mathcal{L}^{-1}$ is the inverse Laplace transform – clathratus Feb 23 '21 at 20:17
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    Yes ${}{}{}{}{}{}{}{}$ – Some Guy Feb 23 '21 at 20:40

2 Answers2

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Contour integration of $(e^{iz}-e^{-a})/(a^2+z^2)$ along the boundary of $|z|<R,0<\arg z<\pi/2$, with $z=ia$ a removable singularity (we assume $a>0$), yields $$\int_0^\infty\frac{\sin x}{a^2+x^2}\,dx=\int_0^\infty\frac{e^{-x}-e^{-a}}{a^2-x^2}\,dx,$$ expressible in terms of the exponential integral function (after doing partial fractions).

metamorphy
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First, we enforce the substitution $x\mapsto ax$. Then, we use the convolution theorem of the Fourier Transform to write

$$\begin{align} I(a)&\equiv \int_0^\infty \frac{\sin(x)}{x^2+a^2}\,dx\\\\ &=\frac1a\int_0^\infty \frac{\sin(ax)}{x^2+1}\,dx\\\\ &=\frac1a \int_{-\infty}^\infty \frac{\sin(ax)}{x^2+1}\,H(x)\,dx\\\\ &=\frac1a\text{Im}\left(\int_{-\infty}^\infty \frac{H(x)}{x^2+1}e^{iax}\,dx\right)\\\\ &=\frac1{2\pi a} \text{Im}\left(\left(\pi e^{-|a|}\right)*\left(\pi \delta(a)+\text{PV}\left(\frac ia\right)\right)\right)\\\\ &=\frac1{2a}\text{PV}\left(\int_{-\infty}^\infty \frac{e^{-|a-a'|}}{a'}\,da'\right)\\\\ &=\frac{e^{-a}\text{Ei}(a)-e^{a}\text{Ei}(-a)}{2a} \end{align}$$

where $\text{Ei}(a)=\int_{-\infty}^a \frac{e^{x}}{x}\,dt$ is the exponential integral, interpreted in the Cauchy Principal Value sense for $a>0$.

Mark Viola
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  • Though being familiar with Fourier Transform and distributions, I am not sure to understand how you obtain (5th line) this convolution... Could you give some detail ? – Jean Marie Feb 23 '21 at 21:12
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    @JeanMarie The [convolution theorem] for the Fourier Transform (for tempered distributions) is discussed HERE. And HERE is a more detailed discussion. The Fourier transform of the Heaviside function, $H$, can be found HERE. – Mark Viola Feb 23 '21 at 22:10
  • Oh, I understand now that you consider the expression on the 4th line as the (imaginary part of the inverse) F.T. of the product $H(x) . \dfrac{1}{1+x^2}$, therefore equal to the conolution of their F.T. .... This astute way (+1) deserves indeed a small explanation, don't you think ?... – Jean Marie Feb 23 '21 at 22:25
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    @jeanmarie Pleased to hear that you understand. I had thought the second sentence before the development would suffice as an explanation. Now, with our comments, the reader is well equipped. Thank you for your very thoughtful comments. I hope you're doing well. Stay safe and healthy my friend. – Mark Viola Feb 23 '21 at 23:39
  • I don't think your answer needs any more improvement (I already upvoted it, as I said you in a comment). – Jean Marie Mar 10 '21 at 21:16
  • @JeanMarie That message was directed explicitly to the OP. – Mark Viola Mar 10 '21 at 23:16
  • I have already upvoted your answer that I find, as it is now, very understandable. Maybe give a reference to the Fourier transform of $H(x)$ – Jean Marie Apr 03 '21 at 07:55
  • @JeanMarie That last comment was directed to the OP. But as a reference to the FT of the Heaviside function, you can find one HERE, and another HERE. – Mark Viola Apr 03 '21 at 14:41
  • @N .S. Hi! It's been a while. I hope you're stayed safe and healthy during the pandemic. If you would, please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to up vote an answer as you see fit. ;-) – Mark Viola Jul 23 '21 at 20:07