Integration, Help!

1. Aug 9, 2012

XtremePhysX

1. The problem statement, all variables and given/known data

Evaluate the following integrals:

2. Relevant equations

(1)$$\int_{0}^{\infty }\frac{cos(x)}{1+(x)^{2}}dx$$
(2)$$\int_{0}^{\infty}e^{-x^2}dx$$
(3)$$\int_{0}^{1}\log\log\left(\frac{1}{x}\right)\,dx$$
(4)$$\int_{0}^{\infty}\frac{x\,dx}{e^x-1}$$
(5)$$\int_{-\infty}^{\infty}\frac{\sin(x)\,dx}{x}$$

3. The attempt at a solution

(2) Let $$I = \int_{0}^{\infty }e^{-x^{2}}dx$$ then we have by the square of an integral:
$$I^{2}=\left ( \int_{0}^{\infty }e^{-x^{2}}dx \right )\left ( \int_{0}^{\infty }e^{-y^{2}}dy \right ) = \int_{0}^{\infty }\int_{0}^{\infty }e^{-x^{2}-y^{2}}dydx$$
Shifting to polar coordinates:
$$\int_{0}^{\frac{\pi }{2}}\int_{0}^{\infty }e^{-r^{2}}rdrd\theta$$
For the interior integral, we use the transformation:
$$u=r^{2}, \frac{du}{2}=rdr$$ to obtain:
$$\frac{1}{2}\int_{0}^{\infty }e^{-u}du = -\frac{1}{2}\left [ e^{-u} \right ]\infty to 0 = \frac{1}{2}$$
Plugging this into our integral, we're left with:
$$I^{2}=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}d\theta = \frac{\pi }{4}$$
By taking the square root of both sides, we have the answer:
$$I=\int_{0}^{\infty }e^{-x^{2}}dx=\frac{\sqrt{\pi }}{2}$$

2. Aug 9, 2012

jackmell

For the first one, I would selfishly resort to contour integration and write:

$$\int_0^{\infty}\frac{\cos(x)}{(x+i)(x-i)}dx=1/2 \int_{-\infty}^{\infty} \frac{\cos(x)}{(x+i)(x-i)}dx$$

then consider the contour integral:

$$\oint \frac{e^{iz}}{(z+i)(z-i)}dz$$

where the contour is the half-disc in the upper half plane, then just use the Residue Theorem but not sure you want to go this way.

3. Aug 9, 2012

XtremePhysX

how would u solve the counter integral, I've never done counter integration.

4. Aug 9, 2012

jackmell

If this is a homework exercise then perhaps you're not expected to know contour integration and need to find another way. I can't think of one at the moment though. Also, really not helpful to you I think for me to try and explain the contour integration if you've not had any Complex Analysis.

5. Aug 9, 2012

XtremePhysX

If counter integration is the better way, then can u show me how to do it that way? I'll learn it.

6. Aug 9, 2012

tiny-tim

Hi XtremePhysX!

(1) looks excessively nasty :yuck:

are you sure it isn't cos-1x on the top?

7. Aug 9, 2012

XtremePhysX

yes I'm sure :) it is cos(x)

8. Aug 9, 2012

jackmell

The third one is one definition of the Euler Mascheroni constant so we would need to show:

$$\int_0^1 \log\left(\log(1/x)\right)dx=-\gamma$$

but I can't think of a way to show this.

9. Aug 9, 2012

gabbagabbahey

Use the derivative of the Gamma function and a the substitution u=log(1/x).

10. Aug 9, 2012

XtremePhysX

Can anyone help me with this integral please:
$$\int^{\frac{\sqrt{3}}{2}}_{\frac{1}{\sqrt{2}}} \frac{1 - 2x^2}{x^5 - x^3}dx$$ using x=sin(u)

11. Aug 9, 2012

XtremePhysX

$$\int_{\frac{\pi }{3}}^{\frac{\pi }{4}}\frac{1-2sin^{2}u}{sin^{5}u-sin^{3}u}cosudu$$

I got up to here.

12. Aug 9, 2012

gabbagabbahey

Why use that substitution?

$$\frac{1-2x^2}{x^5-x^3}=-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)$$

13. Aug 9, 2012

XtremePhysX

Where do you go from here?

14. Aug 9, 2012

XtremePhysX

Do you use a substitution to integrate $$-\frac{1}{x^3} \left (2+\frac{1}{x^2-1}\right)$$

15. Aug 9, 2012

gabbagabbahey

$$\int \left( f(x) + g(x) \right) dx = \int f(x) dx + \int g(x) dx$$

(As long as the integrals converge)

16. Aug 9, 2012

XtremePhysX

Like this:
$$-\frac{1}{x^3}\left ( \int2dx + \int \frac{1}{x^2-1}\right)dx$$

17. Aug 9, 2012

gabbagabbahey

Is $\frac{1}{x^3}$ a constant? If not, how can you pull it out of the integral like that?

18. Aug 9, 2012

XtremePhysX

I don't know, then how do u apply that identity?

19. Aug 9, 2012

Bohrok

Just distribute to the terms in parentheses:$$-\frac{1}{x^3}\left(2 + \frac{1}{x^2 - 1}\right) = -\frac{2}{x^3} - \frac{1}{x^3(x^2 - 1)}$$

20. Aug 10, 2012

XtremePhysX

Can you guys help me with the original integrals please? I kind of know how to do the first, second and third, please help me do the last two.