# Integral calculus by part or with complex numbers

I have to calculate this integral : $$\int_0^{\infty} cos(2xt)exp(-t) dt$$

I try it with integration by part but it's too long, and my teacher should me to use complex numbers because $cos(2xt)=Re(exp(i2xt))$

So now i’m here :

$$\int_0^{\infty} cos(2xt)exp(-t) dt =Re\left( \int_0^{\infty} exp(t(2ix-1)) dt\right)=Re\left[\frac{exp(t(2ix-1))}{2ix-1} dt\right]_0^{\infty}$$

But it seems to be a bit long, I try to expand it but without result can you help me please?

Thank you

• $e^{(2ix-1)t}$ could be better. I guess that you have some typo. By the way, it is not long. Feb 28, 2015 at 16:55
• sorry its an error when I wrote on the site Feb 28, 2015 at 16:59
• If I was given a cent for each of my typo's, I should be a billionaire ! Feb 28, 2015 at 17:18

I would say

$\cos x = \frac{e^{ix}+e^{-ix}}{2}$

$\displaystyle \Rightarrow \int_0^{\infty} cos(2xt)exp(-t) dt=\frac{1}{2}\int_0^{\infty} (e^{2ixt}+e^{-2ixt})e^{-t} dt=$

$=\frac{1}{2}\int_0^{\infty} (e^{-t(-2ix+1)}+e^{-t(2ix+1)}) dt=$

$\displaystyle =\frac{1}{2}\left[\frac{e^{-t(-2ix+1)}}{2ix-1}-\frac{e^{-t(2ix+1)}}{2ix+1}\right]_0^{\infty}=\frac{1}{2}\left(\frac{1}{2ix+1}-\frac{1}{2ix-1}\right)=\frac{1}{4x^2+1}$

• You made it really simple good one Feb 28, 2015 at 17:41

You're pretty much there with the final expression (except for the $\text{d}t$) as far as I can see. Writing out the exponentials might make it a bit more clear: $$\Phi=\int_0^\infty\cos(2xt)e^{-t}\text{d}t = \text{Re}\left[\frac{e^{2ixt}}{(2ix-1)e^t}\right]_{t=0}^\infty.$$ At $t=\infty$ I think we can safely assume that the expression becomes zero (not entirely sure), so that doesn't contribute anything. For $t=0$ the exponentials become $1$, thus we get $$\Phi = \text{Re}\left[-\frac{1}{2ix-1}\right]=\text{Re}\left[-\frac{2ix+1}{(2ix-1)(2ix+1)}\right]=\text{Re}\left[\frac{2ix+1}{4x^2+1}\right].$$ The real part follows as $$\Phi=\frac{1}{4x^2+1}.$$

Another possible approach is the following:

$$\int_{0}^{+\infty}\cos(x)\,e^{-nx}\,dx =\sum_{m\geq 0}\frac{(-1)^m}{(2m)!}\int_{0}^{+\infty}x^{2m}e^{-nx}\,dx=\sum_{m\geq 0}(-1)^m n^{1-2m}=\frac{n}{1+n^2}\tag{1}$$ where we exploited the Taylor series of the cosine function, the integral definition of the $\Gamma$ function and a geometric series. In the same way we get: $$\int_{0}^{+\infty}\sin(x)\, e^{-nx}\,dx = \frac{1}{1+n^2}$$ for every $n>1$ and $$\int_{0}^{+\infty}\frac{\sin(m x)}{m}\,e^{-x}\,dx = \frac{1}{1+m^2}$$ for every $m\in(0,1)$.