Computation real integral with residue theorem

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.
• Are you obliged to use complex analysis and residues ? Dec 6 '21 at 13:21
• @ClaudeLeibovici I think the idea is to use complex analysis and residues but I'm still interested in a different approach Dec 6 '21 at 13:22
• 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 $a<0 \,\, I=-\frac 2 {\pi}$ 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 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 $$\delta \to 0$$.

EDIT

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 $$\delta\to 0$$ $$\int_{-\infty}^\infty \frac{\sin \pi t}{t(1-t)(1+t)}\,\mathrm dt = -\operatorname{Im}(I_{-1} + I_0 + I_1) = 2\pi$$

• 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 $1$? 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.

– Community Bot
Dec 6 '21 at 13:31
• @Community should have more time for a complete answer, this is an essential hint. Dec 6 '21 at 13:33
• "Partical fracture" I like that terminology. Dec 6 '21 at 13:34
• @GEdgar Sorry for Google Translator Quality 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