Improper Integral

• Jun 23rd 2008, 07:10 PM
auslmar
Improper Integral
Hello,

I'm having trouble getting started solving this improper integral:

$
\int\frac{6}{3x^2+5}dx
$
from -infinity to infinity

I don't know how I should approach the problem. Should I use u-substitution, or is there a sure-fire method? I greatly appreciate any tips or suggestions on how to get started.

Austin Martin
• Jun 23rd 2008, 07:28 PM
TheEmptySet
Quote:

Originally Posted by auslmar
Hello,

I'm having trouble getting started solving this improper integral:

$
\int\frac{6}{3x^2+5}dx
$
from -infinity to infinity

I don't know how I should approach the problem. Should I use u-substitution, or is there a sure-fire method? I greatly appreciate any tips or suggestions on how to get started.

Austin Martin

let $x=\left(\frac{\sqrt{5}}{\sqrt{3}} \right)\tan(\theta) \implies dx=\left(\frac{\sqrt{5}}{\sqrt{3}} \right)\sec^2(\theta)d\theta$

When $x=\infty, \theta =\frac{\pi}{2}$ and

$x=-\infty, \theta =-\frac{\pi}{2}$

so

$

\int_{-\infty}^{\infty}\frac{6}{3x^2+5}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\left(\frac{\sqrt{5}}{\sqrt{3}} \right)\sec^2(\theta)d\theta}{3(\left(\frac{\sqrt{ 5}}{\sqrt{3}} \right)\tan(\theta) )^2+5}=\frac{\frac{\sqrt{5}}{\sqrt{3}}}{5}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sec^2(\theta) }{\tan^2(\theta)+1}d\theta
$

$\frac{\sqrt{15}}{15}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}d\theta=\frac{\sqrt{ 15}}{15}\left( \frac{\pi}{2}-\left[ -\frac{\pi}{2}\right]\right)=\frac{\pi \sqrt{15}}{15}$
• Jun 23rd 2008, 07:30 PM
Jhevon
Quote:

Originally Posted by auslmar
Hello,

I'm having trouble getting started solving this improper integral:

$
\int\frac{6}{3x^2+5}dx
$
from -infinity to infinity

I don't know how I should approach the problem. Should I use u-substitution, or is there a sure-fire method? I greatly appreciate any tips or suggestions on how to get started.

Austin Martin

Note that $\int \frac 6{3x^2 + 5}~dx = \frac 65 \int \frac 1{\left( \sqrt{\frac 35}x \right)^2 + 1}~dx$

Now let $u = \sqrt{\frac 35}x$

then $du = \sqrt{\frac 35}~dx \implies \sqrt{\frac 53}~du = dx$

so our integral becomes:

$\frac 65 \cdot \sqrt{\frac 53} \int \frac 1{u^2 + 1}~du$

what's in front of the integral is a constant, the integral itself should look familiar. can you continue?