# integral cosine and exponential

• Jul 13th 2008, 06:00 PM
mathmarco
integral cosine and exponential
I'm trying to solve the following integral

int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx

where a,b and c are real.

thanks a lot

Marco
• Jul 13th 2008, 06:03 PM
Mathstud28
Quote:

Originally Posted by mathmarco
I'm trying to solve the following integral

int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx

where a,b and c are real.

thanks a lot

Marco

$\displaystyle \int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}\cos(cx)~dx$?
• Jul 13th 2008, 09:53 PM
mathmarco
yes, it is exactly that.

I will use \math next time
• Jul 13th 2008, 09:59 PM
Mathstud28
Quote:

Originally Posted by mathmarco
I'm trying to solve the following integral

int_0^{+infintity} 1/(1+a^2*x^2)*exp(-b^2*x^2)*cos(c*x) dx

where a,b and c are real.

thanks a lot

Marco

Well first off I am pretty sure that we must have that $\displaystyle b\in\mathbb{N}$

Otherwise the integral would diverge.
• Jul 13th 2008, 10:22 PM
mathmarco
I though it was not necessary to have $\displaystyle b\in N$
It is not sufficient that $\displaystyle b\in R$

why do you think it will diverge? Does not the following integral converge?
$\displaystyle \int_0^{+\infty} e^{-b^2 x^2} dx$

I'm maybe missing something...
• Jul 13th 2008, 11:00 PM
Mathstud28
Quote:

Originally Posted by mathmarco
I though it was not necessary to have $\displaystyle b\in N$
It is not sufficient that $\displaystyle b\in R$

why do you think it will diverge? Does not the following integral converge?
$\displaystyle \int_0^{+\infty} e^{-b^2 x^2} dx$

I'm maybe missing something...

$\displaystyle \int_0^{\infty}e^{-b^2x^2}~dx=\int_0^{\infty}e^{-(bx)^2}$

Now let $\displaystyle u=bx\Rightarrow\frac{u}{b}=x$

So $\displaystyle dx=\frac{1}{b}$

so we have that when $\displaystyle x\to\infty\Rightarrow{u\to\infty}$

and when

$\displaystyle x\to{0}\Rightarrow{u\to{0}}$

So we have

$\displaystyle \frac{1}{b}\int_0^{\infty}e^{-u^2}~du=\frac{\sqrt{\pi}}{2b}$

Yeah you are right, I misread it at first as $\displaystyle b$ not $\displaystyle b^2$
• Jul 14th 2008, 05:35 PM
mathmarco
ok... good. I'm still trying to solve it but I always end up to nothing good!!
• Jul 14th 2008, 05:52 PM
Mathstud28
Quote:

Originally Posted by mathmarco
ok... good. I'm still trying to solve it but I always end up to nothing good!!

I can do little more than show it converges, are you sure it is integralbe? Have you tried differentiation under the integral sign after a sub of $\displaystyle bx=u$?

$\displaystyle \forall{x}\in\mathbb{R}\quad{-1\leq\cos(cx)\leq{1}}$

So

$\displaystyle -\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\leq\int_0^{\infty}\frac{\cos (cx)\cdot{e^{-b^2x^2}}}{1+a^2x^2}~dx\leq\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx$

Or in other words

$\displaystyle \bigg|\int_0^{\infty}\frac{\cos(cx)e^{-b^2x^2}}{1+a^2x^2}~dx\bigg|\leq\int_0^{\infty}\fra c{e^{-b^2x^2}}{1+a^2x^2}~dx$

and obviously since

$\displaystyle 0\leq\int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\ll\int_0^{\infty}\frac{dx}{1 +a^2x^2}$

Now

$\displaystyle \int_0^{\infty}\frac{dx}{1+(ax)^2}$

Let $\displaystyle ax=\varphi\Rightarrow\frac{\varphi}{a}=x$

So $\displaystyle dx=\frac{1}{a}$

And $\displaystyle \text{As }x\to\infty\Rightarrow\varphi\to\infty$

and $\displaystyle \text{As }x\to{0}\Rightarrow\varphi\to{0}$

So we have

$\displaystyle \frac{1}{a}\int_0^{\infty}\frac{d\varphi}{1+\varph i^2}=\frac{1}{a}\arctan\left(\varphi\right)\bigg|_ 0^{\infty}=\frac{\pi}{2a}$

Therefore $\displaystyle \int_0^{\infty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx$

So we can finally conclude that

$\displaystyle \int_0^{\infty}\frac{e^{-b^2x^2}\cos(cx)}{1+a^2x^2}~dx\quad\text{converges} ~\forall~(a,b,c)\in\mathbb{R}$

Furthermore we see that

$\displaystyle \bigg|\int_0^{\infty}\frac{e^{-b^2x^2}\cos(cx)}{1+a^2x^2}~dx\bigg|\leq\int_0^{\in fty}\frac{e^{-b^2x^2}}{1+a^2x^2}~dx\leq\frac{\pi}{2a}$

Hope this has been some help.