# Thread: Fourier Sine Integral (Complex Analysis)

1. ## Fourier Sine Integral (Complex Analysis)

This question is similar to another one I've posted

Show:
$\int_{-\infty}^{\infty} \frac{\sin{x}}{x^2+4x+5} = \frac{-\pi\sin{2}}{e}$

so I let $f(z) = \frac{\sin{z}}{z^2+4z+5}$
And I used a closed semicircle which contains $z = -2+i$ of radius R.

Let $C_{R} = \{z=Re^{i\theta}: 0\leq\theta\leq\pi\, R<\infty\}$
$C_{+} = \{z=r: -R

I let: $I = \int_{C_{+}} f(z)dz$ and $I_{R} = \int_{C_R} f(z)dz$

By Jordan's Lemma, $I_{R} = 0$ as $R\to\infty$

By the Cauchy Residue Theorem
$
I + I_{R} = I = 2\pi i {Res}_{z=-2+i} f(z)$

Now I use the following theorem to evaluate the theorem
If $p(z)$ is entire and $q(z)$ is analytic everywhere except $z_{0}$ but $q'(z)$ is entire.

The ${Res}_{z=z_0} \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$

However, I can't get the following to evaluate correctly

$I = 2\pi i \frac{\sin{(-2+i)}}{2(-2+i) + 4(-2+i)}$

I am having troubles with $\sin{(-2+i)}$. None of the identities in my text book seem to help.

2. Originally Posted by Haven
This question is similar to another one I've posted

Show:
$\int_{-\infty}^{\infty} \frac{\sin{x}}{x^2+4x+5} = \frac{-\pi\sin{2}}{e}$

so I let $f(z) = \frac{\sin{z}}{z^2+4z+5}$
And I used a closed semicircle which contains $z = -2+i$ of radius R.

Let $C_{R} = \{z=Re^{i\theta}: 0\leq\theta\leq\pi\, R<\infty\}$
$C_{+} = \{z=r: -R

I let: $I = \int_{C_{+}} f(z)dz$ and $I_{R} = \int_{C_R} f(z)dz$

By Jordan's Lemma, $I_{R} = 0$ as $R\to\infty$

By the Cauchy Residue Theorem
$
I + I_{R} = I = 2\pi i {Res}_{z=-2+i} f(z)$

Now I use the following theorem to evaluate the theorem
If $p(z)$ is entire and $q(z)$ is analytic everywhere except $z_{0}$ but $q'(z)$ is entire.

The ${Res}_{z=z_0} \frac{p(z)}{q(z)} = \frac{p(z_0)}{q'(z_0)}$

However, I can't get the following to evaluate correctly

$I = 2\pi i \frac{\sin{(-2+i)}}{2(-2+i) + 4(-2+i)}$

I am having troubles with $\sin{(-2+i)}$. None of the identities in my text book seem to help.

First remember that

$e^{iz}=\cos(z)+i\sin(z)$

So

$\oint \frac{\sin(z)}{z^2+4z+5}dz= Im\left\{\oint\frac{e^{iz}}{(z^2+4z+5)}dz \right\}$

let $g(z)=\frac{e^{iz}}{(z^2+4z+5)}$

Then

$Res(g,z=-2+i)=\frac{e^{-1-2i}}{2i}=-i\frac{\cos(2)}{2e}-\frac{\sin(2)}{2e}$

$\oint \frac{e^{iz}}{z^2+4z+5}dz=2\pi i \left(-i\frac{\cos(2)}{2e}-\frac{\sin(2)}{2e} \right)=\frac{\pi\cos(2)}{e}-i\frac{\pi\sin(2)}{e}$

Now we know that the integral above is the imaginary part of this so

$\int_{-\infty}^{\infty} \frac{\sin(z)}{z^2+4z+5}dz=-\frac{\pi\sin(2)}{e}$