integration - Trigonometric Improper Integral - $int_{-infty}^{infty} frac{xsin ax}{x^2+1}dx$

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 !

Answer

"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 $a$.