I need to compute $$I:=\int_{-\infty}^{\infty} \frac{\sin(ax)}{x(\pi^2-a^2x^2)}dx$$ ($a>0$)

Probably there is a way to compute it with residue theorem.

My thoughts:

  • The singularity at $x=0$ is removable
  • $I=\displaystyle \DeclareMathOperator{Im}{Im} \Im \left( \int_{-\infty}^{\infty} \frac{e^{iax}}{x(\pi^2-a^2x^2)}dx \right)$ . I usually solve those integrals by using the residue theorem to integrate over the upper semicircle (then the integral over the arc goes to 0 as the radius goes to $\infty$ and we are left with I) but this doesn't work since we have non-removable singularities. I also thought about integrating over a circular sector whose border doesn't contain the singularities (the integrand is an even function) but then the integral over the arc does not go to $0$ anymore and I don't get the expression for $I$ anyway.
  • $\begingroup$ Are you obliged to use complex analysis and residues ? $\endgroup$ Dec 6 '21 at 13:21
  • $\begingroup$ @ClaudeLeibovici I think the idea is to use complex analysis and residues but I'm still interested in a different approach $\endgroup$ Dec 6 '21 at 13:22
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    $\begingroup$ Using the decomposition obtained by @Claude Leibovici (and taking $a>0$) $$\int_{-\infty}^{\infty} \frac{sin(ax)}{x(\pi^2-a^2x^2)}dx=-\frac 1 {\pi^2}\Im\int_{-\infty}^{\infty}\Bigg(\frac{1}{2 (t-\pi )}+\frac{1}{2 (t+\pi )}-\frac{1}{t} \Bigg)e^{it}dt$$ $$=-\frac 1 {\pi^2}\Im\int_{-\infty}^{\infty}\Bigg(\frac{e^{i(t+\pi)}}{2 t}+\frac {e^{i(t-\pi)}}{2 t}-\frac {e^{it}}{ t} \Bigg)dt=\frac 2 {\pi^2}\int_{-\infty}^{\infty}\frac{\sin t}{t}dt=\frac 2 {\pi}$$ For $a<0 \,\, I=-\frac 2 {\pi}$ $\endgroup$
    – Svyatoslav
    Dec 6 '21 at 14:55

$$I=\int\frac{\sin (a x)}{x \left(\pi ^2-a^2 x^2\right)}\,dx=\int \frac{\sin (t)}{(t-\pi)t(t+\pi) }\,dt$$ $$\frac{1}{(t-\pi)t(t+\pi) }=\frac 1 {\pi^2}\Bigg[\frac{1}{2 (t-\pi )}+\frac{1}{2 (t+\pi )}-\frac{1}{t} \Bigg]$$

Consider $$\int \frac {\sin(t)}{t+k}\,dt=\int\frac {\sin(u-k)}{u}\,du=\cos (k)\int\frac{ \sin (u)}{u}\,du-\sin (k)\int\frac{ \cos (u)}{u}\,du$$ So now, you just face sine and cosine integrals.

Back to $x$, you will have $$I=\frac 1{2\pi^2}\Big[2 \text{Si}(a x)+\text{Si}(ax-\pi )+\text{Si}(a x+\pi ) \Big]$$ $$J=\int_{-p}^{+p}\frac{\sin (a x)}{x \left(\pi ^2-a^2 x^2\right)}\,dx=\frac 1{\pi^2}\Big[2 \text{Si}(a p)+\text{Si}(ap-\pi )+\text{Si}(a p+\pi ) \Big]\sim \frac 4{\pi^2}\text{Si}(a p)$$ Now, using the asymptotics $$K=\int_{-\infty}^{+\infty}\frac{\sin (a x)}{x \left(\pi ^2-a^2 x^2\right)}\,dx=\frac{2 |a|}{\pi a}$$


Assume without loss of generality that $a$ is positive. Using the change of variable $ax=\pi t$ we get rid of $a$ $$ I = \frac{1}{\pi^2} \int_{-\infty}^\infty \frac{\sin \pi t}{t(1-t)(1+t)}\,\mathrm dt $$ Apply the residue theorem (or Cauchy's theorem) to $\frac{e^{\pi i t}}{t(1-t)(1+t)}$ with the following contour enter image description here

The large semi-circle has radius $\rho$ and the small ones have radius $\delta$. Take the imaginary part both sides, then use dominated convergence theorem as $\rho \to \infty$ and $\delta \to 0$.


For example, the integral around $t=1$ is $$ I_1(\delta) = \int_0^\pi \frac{\exp(\pi i(\overbrace{1+\delta e^{i\theta}}^{t}) )}{(\underbrace{1+\delta e^{i\theta}}_{t}) (\underbrace{-\delta e^{i\theta}}_{1-t}) (\underbrace{2+\delta e^{i\theta}}_{1+t})} \underbrace{(-i \delta e^{i\theta})\,\mathrm d\theta}_{\mathrm dt} =-i\int_0^\pi \frac{ \exp(\pi i \delta e^{i\theta})}{(1+\delta e^{i\theta})(2+\delta e^{i\theta})}\,\mathrm d\theta $$ The integrand is continuous in $(\delta,\theta)$ for $\delta$ small enough, so we can use DCT (or any « continuity under $\int$ » theorem) to get $$ I_1 :=\lim_{\delta\to 0} I_1(\delta) = -i \int_0^\pi \frac{1}{2}\,\mathrm d\theta = -\frac{\pi i}{2} $$ Similarly $I_{-1} = -\frac{\pi i}{2}$ and $I_0=-\pi i$. Hence, as $\rho \to \infty$ and $\delta\to 0$ $$ \int_{-\infty}^\infty \frac{\sin \pi t}{t(1-t)(1+t)}\,\mathrm dt = -\operatorname{Im}(I_{-1} + I_0 + I_1) = 2\pi $$

  • $\begingroup$ I did exactly as you suggested, but I get $\frac{1}{\pi}$ instead of $\frac{1}{\pi}$. I think it is just about computing the 3 integrals on the small semicircles. I get $\frac{1}{2\pi}$+$\frac{1}{\pi}$- $\frac{1}{2\pi}$ I wish the last one was a plus. Do you also get $-\frac{1}{2\pi}$ for the integral around $1$? $\endgroup$ Dec 6 '21 at 14:56

Not a solution

Partial Fraction Decomposition of the denominator is straight forward as

$$ \frac{A}{x} + \frac{B}{\pi-ax} + \frac{C}{\pi+ax} $$

Remains the Sine Integral type. You might consider their asymptotic behavior or win with Residue Calculus.

  • 1
    $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Dec 6 '21 at 13:31
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    $\begingroup$ @Community should have more time for a complete answer, this is an essential hint. $\endgroup$ Dec 6 '21 at 13:33
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    $\begingroup$ "Partical fracture" I like that terminology. $\endgroup$
    – GEdgar
    Dec 6 '21 at 13:34
  • $\begingroup$ @GEdgar Sorry for Google Translator Quality $\endgroup$ Dec 6 '21 at 13:46

As the singularities are removable, I like to translate the contour so that it misses the singularities when we apply $\sin(x)=\frac1{2i}\left(e^{ix}-e^{-ix}\right)$. $$\newcommand{\Res}{\operatorname*{Res}} \begin{align} &\int_{-\infty}^\infty\frac{\sin(ax)}{x(\pi^2-a^2x^2)}\,\mathrm{d}x\\ &=\int_{-i-\infty}^{-i+\infty}\frac{\sin(az)}{z(\pi^2-a^2z^2)}\,\mathrm{d}z\tag1\\ &=\lim_{R\to\infty}\frac1{2i}\left(\int_{U_R}\frac{e^{iaz}}{z(\pi^2-a^2z^2)}\,\mathrm{d}z-\int_{L_R}\frac{e^{-iaz}}{z(\pi^2-a^2z^2)}\,\mathrm{d}z\right)\tag2\\ &=\lim_{R\to\infty}\frac1{2i}\int_{U_R}\frac{e^{iaz}}{z(\pi^2-a^2z^2)}\,\mathrm{d}z\tag3\\ &=\frac{2\pi i}{2i}\left(\Res_{z=-\pi/a}+\Res_{z=0}+\Res_{z=\pi/a}\right)\left(\frac{e^{iaz}}{z(\pi^2-a^2z^2)}\right)\tag4\\ &=\frac{2\pi i}{2i}\left(\frac{-1}{\pi^2-3\pi^2}+\frac1{\pi^2}+\frac{-1}{\pi^2-3\pi^2}\right)\tag5\\ &=\frac2\pi\tag6 \end{align} $$ Explanation:
$(1)$: Because the integrals over the vertical segments vanish as $R\to\infty$,
$\phantom{\text{(1):}}$ the difference is equal to the limit as $R\to\infty$ of the integral over
$\phantom{\text{(1):}}$ $[-R,R]\cup[R,R-i]\cup[R-i,-R-i]\cup[-R-i,-R]$
$\phantom{\text{(1):}}$ which is $0$ since the singularities of the integrand are removable
$(2)$: $U_R=[-R-i,R-i]\cup-i+Re^{+i[0,\pi]}$ (semi-circle in the upper half-plane)
$\phantom{\text{(2):}}$ $L_R=[-R-i,R-i]\cup-i+Re^{-i[0,\pi]}$ (semi-circle in the lower half-plane)
$\phantom{\text{(2):}}$ and the integrals over the semi-circles vanish as $R\to\infty$ since $a\gt0$
$(3)$: there are no singularities inside $L_R$
$(4)$: the integral is $2\pi i$ times the sum of the residues inside $U_R$
$(5)$: evaluate the residues
$(6)$: simplify


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