# [SOLVED]integral = 2pi sum res UHP + pi i sum res real axis

#### dwsmith

##### Well-known member
$$\DeclareMathOperator{\Ima}{Im}$$
$$\DeclareMathOperator{\Res}{Res}$$
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?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
Try the following

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

what do you get ?

#### dwsmith

##### Well-known member
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 $$-\frac{2}{\pi}$$.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
I mean try finding the integral by residues , you will see where your confusion is.

#### dwsmith

##### Well-known member
I mean try finding the integral by residues , you will see where your confusion is.
That didn't help.

#### ZaidAlyafey

##### Well-known member
MHB Math Helper
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.

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