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I am trying to evaluate the definite integral $$\int_0^\infty \frac{\sin ax\ dx}{x^2+b^2}$$ where $a,b>0$. This is a problem on an assignment for a class in complex variables. I understand the mechanics of contour integration, but I am stuck. (I have spoken to four classmates who are also stuck.) I would appreciate a small hint, such as a hint to point me toward the right contour, or a suggestion for how to modify one of my abandoned ideas (below) to make it work. (But please do not work out the whole integral; I want to do that myself.)

Here is what I have tried so far:

  • The integral from $-\infty$ to $\infty$ is zero because the function is odd, so the semicircular contour in the upper half-plane won't do anything.

  • Viewing the integral as $$ \mathrm{Im}\, \int_0^\infty \frac{e^{aiz}dz}{z^2+b^2}, $$ I tried integrating around the contour from $0$ to $R>0$, along the quarter-circle to $iR$, and back down to $0$, with an indentation at the pole at $bi$. The pole is simple so I can get the contribution from the indentation via the residue, and the contribution from the quarter-circle $\to 0$ as $R\to \infty$, but the problem is that I believe the integral along the segment of the imaginary axis from $iR$ to $0$ is imaginary (so contributes to the imaginary part) and blows up around the pole, and generally seems harder to deal with than the original integral.

  • Viewing the integrand with either $\sin az$ or $e^{iaz}$ in the numerator, I tried integrating along the rectangle with vertices $0,R,R+ib,ib$, with an indentation at the vertex $ib$ due to the pole. Again, I don't know how to get control of the integral along the imaginary axis.

  • I tried using integration by parts to get a cosine to come out, but the integrand remains odd so the upper half-plane semicircular contour still does no good.

  • I had a crazy idea that I was unable to carry out. There exists some path thru the origin along which the integrand $$ \frac{e^{aiz}dz}{z^2+b^2} $$ is pure real. This path is the solution to an ordinary differential equation with boundary condition $y(0)=0$. If my calculations are right the equation is $$ y'=\frac{2xy\cos ax-(b^2+x^2-y^2)\sin ax}{2xy\sin ax+(b^2+x^2-y^2)\cos ax}. $$ The idea was to integrate along $0$ to $R$, counterclockwise along the circle with radius $R$ till it hits this curve, and back to the origin along this curve. By construction, the integral along this last part doesn't contribute anything to the imaginary part; meanwhile the part of the circle in the upper half plane $\to 0$ as $R\to \infty$. The curve must lie entirely outside the upper half-plane or this would show that the original integral I'm trying to evaluate is zero (since the imaginary part of $2\pi i$ times the residue is zero), which isn't plausible. Evaluating the desired integral then rests on:
    (a) whether the pole in the lower half-plane ever gets enclosed by this contour (and I'm pretty sure it doesn't), and
    (b) if I can figure out what is going on with the integral along the part of the circle $|z|=R$ in the lower half-plane before it hits the curve; in particular, what its imaginary part is doing asymptotically as $R\to \infty$.

However, I wasn't sure how to pursue these goals any further.

Update: As it turns out, the assignment contained a typo and the integral was supposed to be $$\int_0^\infty \frac{\sin ax\ dx}{x(x^2+b^2)}$$ This makes the integrand even, so it is done easily with the half-circle contour in the upper half-plane. I definitely learned more because of the typo. Thanks all for your answers and comments.

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  • $\begingroup$ Hmm... The integral is not expressible in elementary functions, so simple residues would not do it $\endgroup$ – Sasha Oct 1 '11 at 15:40
  • $\begingroup$ Is this the assignment problem as stated, or what you have reduced it to? If it is as stated, then did it say what form your answer should take? Your integral, even in the special case $a=b=1$ is only expressible in terms of the Exponential integral: mathworld.wolfram.com/ExponentialIntegral.html $\endgroup$ – Ragib Zaman Oct 1 '11 at 15:40
  • $\begingroup$ @ Ragib - verbatim from the assignment, the problem is "Evaluate the following integrals. Justify each step. $\int_0^\infty \left[\frac{dx}{x^2+b^2}\right]\sin ax$ [...and then some other integrals...]." Perhaps there was a mistake? $\endgroup$ – Ben Blum-Smith Oct 1 '11 at 15:57
  • $\begingroup$ @ Ragib or Sasha - if you could give a quick argument that what you're saying is true, I would accept it as an answer. $\endgroup$ – Ben Blum-Smith Oct 1 '11 at 16:15
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You can actually see directly that there is no elementary expression for the integral, using contour integration. For your integral is the same as $$\mathrm{Im}\int_0^\infty {e^{iax} \over x^2 + b^2}\mathrm dx$$ Note that $${1 \over x^2 + b^2} = {1 \over 2ib}\left({1 \over x - ib} - {1 \over x + ib}\right)$$ So it suffices to evaluate the integrals $$\int_0^\infty {e^{iax} \over x \pm bi}\,\mathrm dx$$ These integrals converge despite the denominator having only degree one because the exponential factor modulates it, similar to for $\dfrac{\sin(x)}{x}$. You can use your quarter circle idea on each of these two terms (without even having to worry about indentations)... the upper quarter circle for $+bi$ and the bottom quarter circle for $-bi$. For example you get that the first one is equal to $$\int_0^\infty {e^{-ax} \over x + b}\,\mathrm dx$$ Letting $x = z - b$ this becomes $$e^{ab}\int_b^\infty {e^{-az} \over z}\,\mathrm dz$$ Then letting $z = \dfrac{w}{a}$ this turns into $$e^{ab}\int_{ab}^\infty {e^{-w} \over w}\,\mathrm dw$$ $$= e^{ab}\mathrm{Ei}(-ab)$$ Here $\mathrm{Ei}$ is the exponential function which is defined in terms of the above integral. So unless the exponential functions from the two quarter circles somehow cancel out (which I seriously doubt in view of Sasha's answer), you won't be able to get an elementary expression for the integral.

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  • $\begingroup$ I tried doing this all out and I think I still need the indentation, right? Because the one with $x-ib$ in the denominator still has $e^{aix}$ in the numerator so the integral around the lower quarter-circle doesn't converge, right? But, if I do it around the upper quarter-circle, and I think of the exponential integral in terms of a Cauchy principal value, I think it works; do you agree? $\endgroup$ – Ben Blum-Smith Oct 3 '11 at 3:28
  • $\begingroup$ Ah yes, you are right. Once you get the Cauchy principal value integral, break it up into 0 to $2b$ and $2b$ to infinity pices. The second will be another exponential function involving $E(ab)$, and the first will (after some changes of variables like I did above) become $e^{ab}\int_{0}^{ab} {\sinh(t) \over t}\,dt = e^{ab}Shi(ab)$, which is another nonelementary function $\endgroup$ – Zarrax Oct 3 '11 at 14:39
  • $\begingroup$ +1 Thanks for this answer, it's just what I needed. Btw. since $Ei(x)=P.V.\int_{-\infty}^x$ for $x>0$ (at least this seems to be what Wikipedia says), I think the whole second integral can be done in terms of $Ei$; I'm getting something like $e^{-ab}Ei(ab)$ though I don't have my work with me. Do you think this is right? $\endgroup$ – Ben Blum-Smith Oct 5 '11 at 0:12
  • $\begingroup$ I am not really an expert on special functions and hadn't seen $Ei(x)$ used past the singularity before.. but I have no reason to doubt wikipedia on this. $\endgroup$ – Zarrax Oct 5 '11 at 0:31
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First, the Mellin transform convolution theorem (aka Slater theorem) states that if $\mathcal{M}_s(f) = F(s)$ and $\mathcal{M}_s(g) = G(s)$, then $\int_0^\infty x^{\alpha-1} f(x) g(x) \mathrm{d} x = \mathcal{M}^{-1}\left( F(s) G(s-\alpha) \right)$, where $\mathcal{M}^{-1}$ stands for the inverse Mellin transform.

Now: $$ \mathcal{M}_s(\sin(a x)) = \int_0^\infty x^{s-1} \sin(a x) \mathrm{d} x = \frac{\sqrt{\pi}}{2} \left( \frac{a}{2} \right)^{-s} \frac{\Gamma( \frac{1}{2} + \frac{s}{2} )}{\Gamma(1- \frac{s}{2})} $$ and $$ \mathcal{M}_s\left(\frac{1}{x^2 + b^2}\right) = \frac{b^{s-2}}{2} \Gamma\left(\frac{s}{2}\right)\Gamma\left(1-\frac{s}{2}\right) $$ Therefore, by the Mellin convolution theorem: $$ \begin{eqnarray} \int_0^\infty \frac{\sin(a x)}{x^2+b^2} \mathrm{d} x &=& \mathcal{M}^{-1}\left( \frac{\sqrt{\pi}}{2} \left( \frac{a}{2} \right)^{-s} \frac{\Gamma( \frac{1}{2} + \frac{s}{2} )}{\Gamma(1- \frac{s}{2})} \frac{b^{-s-1}}{2} \Gamma\left(\frac{1}{2}-\frac{s}{2}\right)\Gamma\left(\frac{1}{2}+\frac{s}{2}\right) \right) \\ &=& \frac{\sqrt{\pi}}{2 b} \, G_{1,3}^{2,1}\left(\frac{a^2 b^2}{4}\left| \begin{array}{c} \frac{1}{2} \\ \frac{1}{2},\frac{1}{2},0 \end{array}\right. \right) \\ &=& \frac{\mathrm{Shi}(a b) \cosh (a b)-\mathrm{Chi}(a b) \sinh (a b)}{b} \end{eqnarray} $$ Where $\mathrm{Shi}$ and $\mathrm{Chi}$ stand for hyperbolic sine and cosine integrals, and $ G_{1,3}^{2,1}(z)$ stands for Meijer G-function.

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  • $\begingroup$ As a tiny note: $\mathrm{Shi}$ and $\mathrm{Chi}$ are related to the exponential integral in the same way that $\cosh$ and $\sinh$ are related to $\exp$... $\endgroup$ – J. M. is a poor mathematician Oct 1 '11 at 17:38
  • $\begingroup$ In SWP I got $$\begin{eqnarray*} \int_{0}^{\infty }\dfrac{\sin ax}{x^{2}+b^{2}}dx &=&\dfrac{1}{2b}\pi \sinh ab-\dfrac{1}{2b}\pi \left( \sinh ab\right) \text{signum}\left( a\right) \\ &&+\dfrac{1}{b}\text{Shi}\left( ab\right) \cosh ab-\dfrac{1}{b}\text{Ci}\left( iab\right) \sinh ab \\ &&+\dfrac{1}{2}\dfrac{i}{b}\left( \sinh ab\right) \pi. \end{eqnarray*}$$ $\endgroup$ – Américo Tavares Oct 1 '11 at 19:13
  • $\begingroup$ @AméricoTavares What is SWP ? N.B.: I assumed $a>0$ and $b>0$ per OP. $\endgroup$ – Sasha Oct 1 '11 at 19:45
  • $\begingroup$ SWP is the MacKichan Software Scientific WorkPlace. It has a built-in computer algebra engine. In the version I have (4.10) and where the above result was computed it is Maple. $\endgroup$ – Américo Tavares Oct 1 '11 at 19:55
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    $\begingroup$ @AméricoTavares Mathematica can verify equality of both expression assuming $a>0$ and $b>0$. $\endgroup$ – Sasha Oct 1 '11 at 20:05

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