1. ## Integration tips

Hi

I need to compute the next integral:

$\displaystyle \int_{0}^{\infty} \frac{cos{(tx)}(1- cos{(x)})}{x^2}dx$
$\displaystyle =\int_{0}^{\infty} \frac{cos(tx)}{ x^2} dx - \int_{0}^{\infty} \frac{cos(xt)\cos(x)}{x^2}dx$

but i dont know even where to start ! Any hint how to do this?{

thank you

2. I tried using integration by parts on the first one, and everything seems fine except for the limits...

I want to avoid integrating ont he complex plane but... what do you think?

3. Originally Posted by mabruka
Hi

I need to compute the next integral:

$\displaystyle \int_{0}^{\infty} \frac{cos{(tx)}(1- cos{(x)})}{x^2}dx$
$\displaystyle =\int_{0}^{\infty} \frac{cos(tx)}{ x^2} dx - \int_{0}^{\infty} \frac{cos(xt)\cos(x)}{x^2}dx$

but i dont know even where to start ! Any hint how to do this?{

thank you
Hint:

Spoiler:
For the second one use the product-to-sum identity to make it into two similar cases of the first integral. For that one use the fact that $\displaystyle \int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy$

4. Thank you, im gonna work in it right now

5. Originally Posted by Drexel28
Hint:

For the second one use the product-to-sum identity to make it into two similar cases of the first integral. For that one use the fact that $\displaystyle \int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy$
Sorry, i dont understand the fact. What fact about that integral? The only think i can think of is its relation with the gamma function

$\displaystyle \int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy = \frac{1}{x^2}\int_0^{\infty}u\cdot e^{-u}\text{ }du = \frac{ \Gamma (2)}{x^2} = \frac{ 1}{x^2}$

6. Guess im gonna go over the complex field to integrate that last integral!

Hope it works just fine

Thnx for the help

7. Originally Posted by mabruka
Sorry, i dont understand the fact. What fact about that integral? The only think i can think of is its relation with the gamma function

$\displaystyle \int_0^{\infty}y\cdot e^{-y\cdot x}\text{ }dy = \frac{1}{x^2}\int_0^{\infty}u\cdot e^{-u}\text{ }du = \frac{ \Gamma (2)}{x^2} = \frac{ 1}{x^2}$
$\displaystyle \int_0^{\infty}\frac{\cos(tx)}{x^2}dx=\int_0^{\inf ty}\int_0^{\infty}ye^{-yx}\cos(tx)dydx$ switch the order of integration.

8. Nice try but that integral diverges.

And you'll need a justification to swap the integration order.

9. After working with it a bit i found out

$\displaystyle \int_0^{\infty}\frac{\cos(tx)}{x^2}dx=\int_0^{\inf ty}\int_0^{\infty}ye^{-yx}\cos(tx)dydx = \int_{0}^{ \infty}\frac{y^2}{y^2+t^2} dy$

which as Krizalid pointed out it is not finite.

The problem here is we can not use Fubini's theorem because the function

$\displaystyle f(x,y)= ye^{-yx}\cos(tx)$ is not integrable.

So the only way i can think of doing it is over the complex field.