# Integrating using residue theorem from minus infinity to infinity when there are zeroes on the real line

I'm practising some residue calculations, and there's a few questions that require me to integrate from $-\infty$ to $\infty$. Usually this is fine, I do it over the big semicircle and let R go to infinity and sum the residues in the upper half plane. But in this question the $x$ term has a zero at zero.

$$\int^\infty_{-\infty} \frac{sin(mx)}{x(a^2 +x^2)} dx \ \text{for} \ a>0$$

I thought about using partial fractions but I can't get it to split. Can anyone offer help or a diagram of where to begin?

Thanks

## 2 Answers

$$\frac{\sin(mx)}{x(a^2+x^2)}=\sin(mx)\frac{1}{x(a^2+x^2)}=\left(\frac{\sin(mx)}{a^2x}-\frac{x\sin(mx)}{a^2(a^2+x^2)}\right)$$

But more simply, treating $z=x+0i$

$$\frac{\sin(mz)}{z(a^2+z^2)}=\frac{\sin(mz)}{z(z+ai)(z-ai)}$$

which has poles of order $1$ at $z=\pm ai$

• wait so what's the reason this doesn't have a pole at z=0? – alsowarwickstudent May 7 '18 at 18:39
• Because $\sin(0)=0$ which makes $z=0$ a removable pole, or rather $\displaystyle\lim_{z\to0}\frac{\sin(mz)}{z}=1$ – AEngineer May 7 '18 at 20:03
• Oh brilliant thank you! – alsowarwickstudent May 8 '18 at 17:51

# Hint:

$$\int_{-\infty}^{\infty}{\sin\left(mx\right) \over x\left(x^{2} + a^{2}\right)}\,\mathrm{d}x = \mathrm{sgn}\left(m\right)\Im\int_{-\infty}^{\infty}{\mathrm{e}^{\mathrm{i}\left\vert m\right\vert x} - 1\over x\left(x^{2} + a^{2}\right)}\,\mathrm{d}x$$