Integral = 2pi sum res UHP + pi i sum res real axis

[Dustinsfl]

\(\DeclareMathOperator{\Ima}{Im}\)
\(\DeclareMathOperator{\Res}{Res}\)
Given
\[
\Ima\left[\int_{-\infty}^{\infty}\frac{e^{iz}}{z(\pi^2 - z^2)}dz\right].
\]
I know the integral is equal to
\[
2\pi i\sum_{\text{UHP}}\Res(f(z); z_j) + \pi i\sum_{\mathbb{R}\text{ axis}}\Res(f(z); z_k).
\]
However, the poles are \(z = 0\) and \(z = \pm\pi\) which are all on the real axis so we just have the sum on the real axis.
\[
\pi i\sum\lim_{z\to z_j}(z - z_j)\frac{e^{iz}}{z(\pi^2 - z^2)} =
\pi i\left[\frac{1}{\pi^2} + \frac{1}{2\pi^2} - \frac{1}{2\pi^2}\right] = \frac{i}{\pi}
\]
However, the solution is \(\frac{2}{\pi}\). What is wrong?

 

[alyafey22]

Try the following

\(\displaystyle \int^{\infty}_{-\infty}\frac{e^{iz}}{z(z^2-\pi^2)}\, dz\)

what do you get ?

 

[Dustinsfl]

Try the following

\(\displaystyle \int^{\infty}_{-\infty}\frac{e^{iz}}{z(z^2-\pi^2)}\, dz\)

what do you get ?

If you do that, you need to pick up a negative sign out front then though.

In Mathematica, that integral is \(-\frac{2}{\pi}\).

 

[alyafey22]

I mean try finding the integral by residues , you will see where your confusion is.

 

[Dustinsfl]

I mean try finding the integral by residues , you will see where your confusion is.

That didn't help.

 

[alyafey22]

By residues we have

\(\displaystyle \lim_{z\to \pi }(z-\pi) \frac{e^{iz}}{z(\pi^2-z^2)}=\lim_{z\to \pi } \frac{e^{iz}}{-z(z+\pi)}=\frac{1}{2\pi^2}\)

Also we have

\(\displaystyle \lim_{z\to -\pi }(z+\pi) \frac{e^{iz}}{z(\pi^2-z^2)}=\lim_{z\to - \pi } \frac{e^{i\pi}}{z(\pi-z)}=\frac{1}{2\pi^2}\)

So they don't cancel.