Find integral for this

**bcvw85** · May 28th 2008, 11:08 AM

Hi i'm lookin for the integral of t*exp(-abs(t))sin(wt) dt with limits at +infinity and -infinity

Anyone available to help?

**Leslon** · May 28th 2008, 11:29 AM

Well, I have a solution but no steps..

I hope this will help you though.

$\frac{4w}{(1+w^2)^2}$
Greets,Leslon

**flyingsquirrel** · May 28th 2008, 11:46 AM

Hi

Originally Posted by bcvw85

Hi i'm lookin for the integral of t*exp(-abs(t))sin(wt) dt with limits at +infinity and -infinity

First, let's get rid off the absolute value. As $t\mapsto t\sin(\omega t)\exp(-|t|)$ is an even function,

$I=\int_{-\infty}^{\infty}t\sin(\omega t)\exp(-|t|)\,\mathrm{d}t=2\int_0^{\infty}t\sin(\omega t)\exp(-t)\,\mathrm{d}t$

Then, one can use the expression of the sine in terms of exponential functions : $\sin(\omega t)=\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i}$ which gives

$\begin{aligned} I&=\int_0^{\infty}t\left(\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i}\right)\exp(-t)\,\mathrm{d}t\\ &=\int_0^{\infty}t\left(\frac{\exp((i\omega-1) t)-\exp(-(1+i\omega) t)}{2i}\right)\,\mathrm{d}t\end{aligned}$

Using integration by parts one can get rid off the $t$ which is outside the exponential functions.

**CaptainBlack** · May 28th 2008, 11:50 AM

Originally Posted by flyingsquirrel

Hi

First, let's get rid off the absolute value. As $t\mapsto t\sin(\omega t)\exp(-|t|)$ is an even function,

$I=\int_{-\infty}^{\infty}t\sin(\omega t)\exp(-|t|)\,\mathrm{d}t=2\int_0^{\infty}t\sin(\omega t)\exp(-t)\,\mathrm{d}t$

Then, one can use the expression of the sine in terms of exponential functions : $\sin(\omega t)=\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i}$ which gives

$\begin{aligned} I&=\int_0^{\infty}t\left(\frac{\exp(i\omega t)-\exp(-i\omega t)}{2i}\right)\exp(-t)\,\mathrm{d}t\\ &=\int_0^{\infty}t\left(\frac{\exp((i\omega-1) t)-\exp(-(1+i\omega) t)}{2i}\right)\,\mathrm{d}t\end{aligned}$

Using integration by parts one can get rid off the $t$ which is outside the exponential functions.

You are making hard work of this, see the other thread

RonL

**Chris L T521** · May 28th 2008, 11:59 AM

Originally Posted by bcvw85

Hi i'm lookin for the integral of t*exp(-abs(t))sin(wt) dt with limits at +infinity and -infinity

Anyone available to help?

$\int_{-\infty}^{\infty}te^{-|t|}\sin(wt)\,dt$

Write it as two separate integrals:

$\int_{-\infty}^{0}te^{t}\sin(wt)\,dt+\int_{0}^{\infty}te^ {-t}\sin(wt)\,dt$

Apply integration by parts:

let $u=t$ and $dv=e^{t}\sin(wt)\,dt$

Therefore, $\,du=\,dt$ and $v=\frac{1}{w^2+1}\cdot\left(e^t\sin(wt)-e^t\cos(wt)\right)$ (you get v by using integration by parts...again...

Plugging into the integration by parts formula:

$\left.\left[\frac{te^t}{w^2+1}\cdot\left(\sin(wt)-\cos(wt)\right)\right]\right|_{-\infty}^{0}-\frac{1}{w^2+1}\int_{-\infty}^{0}\left(e^t\sin(wt)-e^t\cos(wt)\right)\,dt$

Before we move on, note that $\left.\left[\frac{te^t}{w^2+1}\cdot\left(\sin(wt)-\cos(wt)\right)\right]\right|_{-\infty}^{0}$ converges to 0.

Thus, we have left:

$-\frac{1}{w^2+1}\int_{-\infty}^{0}\left(e^t\sin(wt)-e^t\cos(wt)\right)\,dt$

Apply integration by parts (again) for these two integals:

$-\frac{1}{w^2+1}\int_{-\infty}^{0}e^t\sin(wt)+\frac{1}{w^2+1}\int_{-\infty}^{0}e^t\cos(wt)\,dt$

I will leave steps out, hoping that you can do it on your own.

Now, the integrals should become $\left.\left[\frac{2we^t\cos(wt)}{(w^2+1)^2}\right]\right|_{-\infty}^{0}-\left.\left[\frac{(w^2-1)e^t\sin(wt)}{(w^2+1)^2}\right]\right|_{-\infty}^{0}$

$\left[\frac{2w}{(w^2+1)^2}-0\right]-\left[0-0\right]=\color{red}\boxed{\frac{2w}{(w^2+1)^2}}$

Since we only evaluated half of the integral, double it to get the answer:

$\therefore \int_{-\infty}^{\infty}te^{-|t|}\sin(wt)\,dt=\color{red}\boxed{\frac{4w}{(w^2+ 1)^2}}$

Hope that this made sense!!

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