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