# Taking down an integral

• Feb 4th 2010, 05:32 PM
arbolis
Taking down an integral
I was wondering which method you'd use to take down the following integral: $\displaystyle \int _0^{\infty} \frac{dx}{(d^2+x^2)^2}$.
I need it to be "killed", so if any of you, inte-killers, are interested, please have fun.
• Feb 4th 2010, 06:34 PM
Soroban
Hello, arbolis!

Quote:

$\displaystyle \int _0^{\infty} \frac{dx}{(a^2+x^2)^2}$.

Let: .$\displaystyle x \:=\:a\tan\theta \quad\Rightarrow\quad dx \:=\:a\sec^2\!\theta\,d\theta$
. . Then: .$\displaystyle a^2+x^2 \:=\:a^2\sec^2\!\theta$

Substitute: .$\displaystyle \int \frac{a\sec^2\!\theta\,d\theta}{(a^2\sec^2\!\theta )^2} \;=\; \frac{1}{a^3}\int\frac{dx}{\sec^2\!\theta} \;=\;\frac{1}{a^3}\int\cos^2\!\theta\,dx \;=\;\frac{1}{2a^3}\int(1 + \cos2\theta)\,dx$

. . . . . . . . $\displaystyle =\;\frac{1}{2a^3}\bigg(\theta + \frac{1}{2}\sin 2\theta \bigg) + C \;=\;\frac{1}{2a^3}\bigg(\theta + \sin\theta\cos\theta\bigg) + C$

Back-substitute: .$\displaystyle \tan\theta \,=\,\frac{x}{a} \quad\Rightarrow\quad \sin\theta \,=\,\frac{x}{\sqrt{a^2+x^2}} \qquad\Rightarrow\quad \cos\theta \,=\,\frac{a}{\sqrt{a^2+x^2}}$

. . and we have: .$\displaystyle \frac{1}{2a^3}\bigg(\arctan\frac{x}{a} + \frac{ax}{a^2+x^2}\bigg) + C$

Now we evaluate from 0 to $\displaystyle \infty$

When we substitute $\displaystyle x = 0$, we get 0.

We don't substitute $\displaystyle \infty$. we have to sneak up on it.

We substitute $\displaystyle x = b$, then take limit as $\displaystyle b$ gets infinite.

We have: .$\displaystyle \lim_{b\to\infty}\frac{1}{2a^3}\bigg(\arctan\frac{ b}{a} + \frac{ab}{a^2+b^2}\bigg)$

. . The first limit is: .$\displaystyle \lim_{b\to\infty}\,\arctan\frac{b}{a} \;=\;\arctan\infty \;=\;\frac{\pi}{2}$

. . The second limit is: .$\displaystyle \lim_{b\to\infty}\frac{ab}{a^2+b^2}$

. . Divide top and bottom by $\displaystyle b^2\!:$ .$\displaystyle \lim_{b\to\infty}\,\frac{\frac{a}{b}}{\frac{a^2}{b ^2} + 1} \;=\;\frac{0}{0+1} \:=\:0$

Therefore, we have: .$\displaystyle \frac{1}{2a^3}\bigg(\frac{\pi}{2} + 0\bigg) \;=\;\frac{\pi}{4a^3}$

Am I close?

• Feb 4th 2010, 06:34 PM
Krizalid
okay then, let's kill it.

$\displaystyle \int_{0}^{\infty }{\frac{dx}{\left( d^{2}+x^{2} \right)^{2}}}=\frac{1}{d^{2}}\left( \int_{0}^{\infty }{\frac{dx}{d^{2}+x^{2}}}-\int_{0}^{\infty }{\frac{x^{2}}{\left( d^{2}+x^{2} \right)^{2}}} \right)\,dx,$ well actually, i'll let you to kill it, 'cause you just need to integrate by parts the last integral and you're done.
• Feb 5th 2010, 11:36 AM
arbolis
Thanks to both! That was helpful.
Having the result, I realize I made an error setting up this integral. It came from an attempt to find the electric field of a semi-infinite charged line. I should have reached $\displaystyle \frac{1}{d^2}$.