# Math Help - Taking down an integral

1. ## Taking down an integral

I was wondering which method you'd use to take down the following integral: $\int _0^{\infty} \frac{dx}{(d^2+x^2)^2}$.
I need it to be "killed", so if any of you, inte-killers, are interested, please have fun.

2. Hello, arbolis!

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

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

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

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

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

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

Now we evaluate from 0 to $\infty$

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

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

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

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

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

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

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

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

Am I close?

3. okay then, let's kill it.

$\int_{0}^{\infty }{\frac{dx}{\left( d^{2}+x^{2} \right)^{2}}}=\frac{1}{d^{2}}\left( \int_{0}^{\infty }{\frac{dx}{d^{2}+x^{2}}}-\int_{0}^{\infty }{\frac{x^{2}}{\left( d^{2}+x^{2} \right)^{2}}} \right)\,dx,$ well actually, i'll let you to kill it, 'cause you just need to integrate by parts the last integral and you're done.

4. 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 $\frac{1}{d^2}$.