Exercise :
If a<0, show that :
$$\int_{-\infty}^{\infty} \frac{x\sin ax}{x^2+1}dx = -\pi e^a$$
Attempt :
$$\int_{-\infty}^{\infty} \frac{x\sin ax}{x^2+1}dx= \Im \Bigg(\int_{-\infty}^{\infty} \frac{xe^{iax}}{x^2+1}dx\Bigg)$$
It is :
$$f(z) = \frac{z}{z^2+1}$$
Because :
$$P(z)=x$$
$$Q(z) = x^2 +1$$
It is :
$$Degree\{ Q(z)\} = Degree\{ P(z)\} +1$$
So, the needed inequality : $Degree\{ Q(z)\} \geq Degree\{ P(z)\} +1$, is satisfied.
In this case :
$$\int_{-\infty}^{\infty} \frac{xe^{iax}}{x^2+1}dx = 2\pi i Res\bigg(\frac{ze^{iaz}}{z^2+1}, z_1\bigg)$$
where $z_1$ is the individual singular point of $f(z) = \frac{z}{z^2+1}$, that resides in the half-upper plane.
Because $z^2+1 = 0 \Leftrightarrow z_{1,2} = \{i,-i\}$, are simple roots of the equation, which means there are simple poles of $f(z)$.
The root $z_1 = i$ resides in the upper-half plane, so :
$$\int_{-\infty}^{\infty} \frac{xe^{iax}}{x^2+1}dx = 2\pi i Res\bigg(\frac{ze^{iaz}}{z^2+1}, i\bigg)$$
Since it is : $z^2+1|_{z=i} = 0$, $2z |_{z=i} \neq 0$ and $ze^{iaz} |_{z=i} \neq 0$ the residue is :
$$Res\bigg(\frac{ze^{iaz}}{z^2+1}, i\bigg) = \frac{ze^{iaz}}{2z}\bigg|_{z=i} = \frac{e^{-a}}{2}$$
So :
$$2\pi i \bigg(\frac{e^{-a}}{2}\bigg)= i \pi e^{-a} \Rightarrow \Im \{i \pi e^{-a}\}= \pi e^{-a}$$
So I get that :
$$\int_{-\infty}^{\infty} \frac{x\sin ax}{x^2+1}dx = \pi e^{-a}$$
But this is not equal to :
$$\int_{-\infty}^{\infty} \frac{x\sin ax}{x^2+1}dx = -\pi e^a$$
which is what I must show.
So, where is my mistake ? Can you help me figure out what I've done wrong ? Please assist me, because I got stuck with finding my mistake.
Thanks in advance !
"So, the needed inequality ... is satisfied".
To do what?
Note that $a<0$, so you cannot integrate along the boundary of a half moon in the upper half plane (which I guess is what lies behind your method) since the exponential part will become big.
I suggest that you either
1) integrate in the lower half plane (and thus calculate the residue at $z=-i$), or
2) note that your method actually works for $a>0$, and use that your integral is odd in $a$.