# Trigonometric Improper Integral - $\int_{-\infty}^{\infty} \frac{x\sin ax}{x^2+1}dx$

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.

"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 $a$.

• You are correct, yes, I should be integrating in the lower-half plane. BUT even if integrating in the lower-half plane and calculating the residue at $z=-i$, I will get $\pi e^a$ but I will be missing the minus $-$ at $-\pi e^a$. Where does that minus come from ? – Rebellos Jun 19 '17 at 11:00
• My (bad) sense of humor, maybe. If you integrate along the boundary of the half moon in the lower half plane, you have to be careful with the orientation of the curve. If you do it wrongly, you might get the wrong sign. Please check your calculations. – mickep Jun 19 '17 at 11:05
• Oh yes, the orientation will be the opposite of the half-upper moon, so it will be minus the integral. My silly question though, it's okay, I've just handled like 10 exercises with integrations on the upper-half moon so I wasn't paying attention to the signs. I get it all now, much thanks ! – Rebellos Jun 19 '17 at 11:09