Integrals: Homework Help & Solutions

[XtremePhysX]

Homework Statement

Evaluate the following integrals:

Homework 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)$$\int_{-\infty}^{\infty}\frac{\sin(x)\,dx}{x}$$

The Attempt at a Solution

I know how to do the second one, so please help me with the rest:

(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:
$$I=\int_{0}^{\infty }e^{-x^{2}}dx=\frac{\sqrt{\pi }}{2}$$

 

[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.

 

[XtremePhysX]

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

 

[jackmell]

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

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.

 

[XtremePhysX]

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.

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

 

[tiny-tim]

Hi XtremePhysX!

(1) looks excessively nasty

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

 

[XtremePhysX]

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

 

[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.

 

[gabbagabbahey]

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.

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

 

[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)

please, please.

 

[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.

 

[gabbagabbahey]

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)

please, please.

Why use that substitution?

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

 

[XtremePhysX]

Why use that substitution?

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

Where do you go from here?

 

[XtremePhysX]

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

 

[gabbagabbahey]

Where do you go from here?

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

(As long as the integrals converge)

 

[XtremePhysX]

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

 

[gabbagabbahey]

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

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

 

[XtremePhysX]

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

 

[Bohrok]

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

 

[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.

 

[Millennial]

For the fourth, use the identity
$$\int_{0}^{\infty}e^{-nx}x^{s-1}dx=\frac{\Gamma(s)}{n^s}$$
along with the monotone convergence theorem.
The fifth can be obtained by considering the complex valued integral
$$\int_{C}\frac{e^{iz}}{z}\,dz$$
on a semicircular contour.

 

[XtremePhysX]

For the fourth, use the identity
$$\int_{0}^{\infty}e^{-nx}x^{s-1}dx=\frac{\Gamma(s)}{n^s}$$
along with the monotone convergence theorem.

Thank you mill :)

 

[gabbagabbahey]

The fifth can also be done by looking at $\frac{d}{da}\int_{-\infty}^{\infty}\frac{e^{iax}}{x}dx$