- #1

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## Homework Statement

Evaluate the following integrals:

## Homework Equations

(1)[tex]\int_{0}^{\infty }\frac{cos(x)}{1+(x)^{2}}dx[/tex]

(2)[tex]\int_{0}^{\infty}e^{-x^2}dx[/tex]

(3)[tex]\int_{0}^{1}\log\log\left(\frac{1}{x}\right)\,dx[/tex]

(4)[tex]\int_{0}^{\infty}\frac{x\,dx}{e^x-1}[/tex]

(5)[tex]\int_{-\infty}^{\infty}\frac{\sin(x)\,dx}{x}[/tex]

## The Attempt at a Solution

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

(2) Let [tex]I = \int_{0}^{\infty }e^{-x^{2}}dx[/tex] then we have by the square of an integral:

[tex]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[/tex]

Shifting to polar coordinates:

[tex]\int_{0}^{\frac{\pi }{2}}\int_{0}^{\infty }e^{-r^{2}}rdrd\theta[/tex]

For the interior integral, we use the transformation:

[tex]u=r^{2}, \frac{du}{2}=rdr[/tex] to obtain:

[tex]\frac{1}{2}\int_{0}^{\infty }e^{-u}du = -\frac{1}{2}\left [ e^{-u} \right ]\infty to 0 = \frac{1}{2}[/tex]

Plugging this into our integral, we're left with:

[tex]I^{2}=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}d\theta = \frac{\pi }{4}[/tex]

By taking the square root of both sides, we have the answer:

[tex]I=\int_{0}^{\infty }e^{-x^{2}}dx=\frac{\sqrt{\pi }}{2}[/tex]