# Evaluate $\int_{-\infty}^{\infty}\frac{z \sin (z)}{\left(z^2+1\right) \left(z^2+2\right)} \mathrm{d}z$

I need help integrating $$\int_{-\infty}^{\infty}\frac{z \sin (z)}{\left(z^2+1\right) \left(z^2+2\right)} \mathrm{d}z$$

I calculated the integral over the closed upper half circle in the complex plane which is $$\pi(\sinh(\sqrt2) - \sinh(1))$$.

Then if I calculate the integral over $$z = R e^{i\phi}$$ for $$\phi \in [0, \pi]$$, and do the limit $$R$$ to $$\infty$$, the integral must be zero. And it follows that

$$\int_{-\infty}^{\infty}\frac{z \sin (z)}{\left(z^2+1\right) \left(z^2+2\right)}\mathrm{d}z = \pi(\sinh(\sqrt2) - \sinh(1))$$

But Wolfram Alpha says this is false.

What did I do wrong?

If you let $\displaystyle f(z) = \frac{z \sin z}{(z^{2}+1)(z^{2}+2)}$, the integral doesn't go to zero along the upper half of $|z|=R$ as $R \to \infty$.

Instead let $\displaystyle f(z) = \frac{z e^{iz}}{(z^{2}+1)(z^{2}+2)}$.

By Jordan's lemma, the integral along the upper half of $|z|=R$ goes to zero as $R \to \infty$.

http://en.wikipedia.org/wiki/Jordan%27s_lemma

There are two simple poles inside of the closed contour at $z=i$ and $z=i \sqrt{2}$.

So $$\int_{-\infty}^{\infty} \frac{x e^{ix}}{(x^{2}+1)(x^{2}+2)} \ dx = 2 \pi i \Big( \text{Res}[f(z),i] + \text{Res}[f(z),i \sqrt{2} \Big)$$

where

$$\text{Res}[f(z),i] = \lim_{z \to i} \ (z-i) \frac{z e^{iz}}{(z^{2}+1)(z^{2}+2)} = \frac{ie^{-1}}{(2i)(1)} = \frac{1}{2e}$$

and

$$\text{Res}[f(z),i\sqrt{2}] = \lim_{z \to i \sqrt{2}} \ (z-i\sqrt{2}) \frac{z e^{iz}}{(z^{2}+1)(z^{2}+2)} = \frac{i \sqrt{2} e^{-\sqrt{2}}}{(-1)(2 i \sqrt{2})} = - \frac{e^{-\sqrt{2}}}{2}$$

Then

$$\int_{-\infty}^{\infty}\frac{x e^{ix}}{(x^{2}+1)(x^{2}+2)} \ dx = 2 \pi i \left( \frac{1}{2e} - \frac{e^{-\sqrt{2}}}{2} \right)= i \pi\left(\frac{1}{e} - e^{-\sqrt{2}} \right)$$

Now just equate the imaginary parts on both sides of the equation.