How to show $\int^{\infty}_{-\infty}\frac{\sin(ax)}{x(x^2+1)}dx=\pi(1-e^{-a})$? ($a\ge0$)

Question

$$\int^{\infty}_{-\infty}\frac{\sin(ax)}{x(x^2+1)}dx=\pi(1-e^{-a}), \ a\ge0$$

I tried to solve but came up with $\pi(2-e^{-a}) $. Could you tell me where did I do the mistake?

if $x=z$ then $dz=dx$

$$\int_\gamma \frac{e^{iaz}}{z(z^2+1)}\quad and\quad z(z+i)(z-i)=0\quad \rightarrow z=0,z=\pm i$$

for $ z=0$ $$Res(f,0)=\lim_{z\to 0}\frac{z.e^{iaz}}{z(z^2+1)}=1$$

for $z=1$

$$Res(f,i)=\lim_{z\to i}\frac{(z-i).e^{iaz}}{z(z+i)(z-i)}=\lim_{z\to i}\frac{e^{iaz}}{z(z+i)}=\frac{e^{-a}}{-2}$$

$$\int_\gamma \frac{e^{iaz}}{z(z^2+1)}=\int_{-R}^{R}\frac{e^{iax}}{x(x^2+1)}dx+\int_\gamma \frac{e^{iaz}}{z(z^2+1)}=\pi i(1-\frac{e^{-a}}{-2})$$

$$\int_{-R}^{R}\frac{e^{iax}}{x(x^2+1)}=\int^{R}_{-R}\frac{\cos(ax)}{x(x^2+1)}dx+i\int_{-R}^R \frac{\sin(ax)}{x(x^2+1)}=i(2\pi - 2\pi \frac{e^{-a}}{2})$$

$$\rightarrow \int_{-R}^R\frac{\sin(ax)}{x(x^2+1)}=2\pi -2\pi \frac{e^{-a}}{2}=\pi(2-e^{-a}) $$

Answer 1

The integral depends on the value of $a$, which I assume is a real number.

Assuming that $a\ge 0$, You can use this kind of contour to integrate $f(z) = \dfrac{e^{iaz}}{z(z^2+1)}$. The point is that you have to go around the singularity at $z=0$. (An alternative is to follow Did's suggestion).

Using this contour, you obtain:

$$2\pi i Res(f,i) = \left(\int_R - \int_\epsilon\right) + \left(\int_r^R f(x) dx + \int_{-R}^{-r} f(x) dx\right),$$

$\int_R$ and $\int_\epsilon$ both represent integral of $f$ over the respective semicircles. A quick computation shows that $R\to \infty$ makes $\int_R\to 0$ and $\epsilon \to 0$ makes $\int_\epsilon\to i\pi$.

Therefore,

$$2\pi i Res(f,i) = - i\pi + \int_{-\infty}^\infty f(x) dx$$

Computing the residue and solving for the integral gives:

$$ \int \dfrac{\sin(ax)}{x(x^2+1)} dx = \pi (1-e^{-a})$$

For $a<0$, you can use a similar contour, but it has to be upside down from the one used above in order for the big $R$ contour to vanish as $R\to \infty$. The answer comes out to be $\pi (e^a -1)$.

Answer 2

I will assume $a > 0$. Call the integral $I(a)$. Differentiate w.r.t. $a$, $$ I'(a) = \int_{-\infty}^{\infty} \frac{\cos(a x)}{1+x^2} \ \mathrm{d}x = 2\int_{0}^{\infty}\frac{\cos(a x)}{1+x^2} \ \mathrm{d}x,$$ then note that $$\mathcal{L}(I'(a)) = 2\int_{0}^{\infty}\int_{0}^{\infty} \frac{\cos(a x)}{1+x^2}\exp(-as) \ \mathrm{d}a \ \mathrm{d}x,$$ $$\mathcal{L}(I'(a)) = 2\int_{0}^{\infty} \frac{s}{(x^2+s^2)(1+x^2)} \ \mathrm{d}x.$$ Using partial fractions and integrating, $$\mathcal{L}(I'(a)) = \frac{\pi}{s+1},$$ taking the inverse transform gives $$ I'(a) = \pi \mathrm{e}^{-a}, $$ integrate to obtain $$ I(a) = -\pi \mathrm{e}^{-a} + c, $$ and using $I(0) = 0$, we see that $c=\pi$ and $$ I(a) = \pi(1-\mathrm{e}^{-a}). $$