How do I integrate \frac{s^3}{(s^2 + a^2)^3} from 0 to infinity?

[fluxions]

Ok, so I need to evaluate this integral: $$\int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds$$
It seems evident that a substitution ought to do the trick. Let's try $u = s^2 + a^2$, so $du = 2sds$. Hence

$$\int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds <br /> = \frac{1}{2} \int_{a^2}^{\infty} \frac{u - a^2}{u^3} du<br /> = \frac{1}{2} \int_{a^2}^{\infty} \frac{1}{u^2} - \frac{a^2}{u^3} ds<br /> = \frac{1}{2} \left(-\frac{1}{u} + \frac{1}{2} \frac{a^2}{u^2}\right)\right|_{a^2}^{\infty}<br /> = \frac{1}{2} \left(\frac{1}{a^2} - \frac{1}{2}\right)$$

This isn't right. In fact I know the answer must be [itex]\frac{1}{4a^2} [/tex]. (I came acorss this integral doing Griffiths E&M problem 8.4; for physical reasons the integral MUST have the value I indicated.) Indeed that's exactly what WolframAlpha gives; see <a href="http://www.wolframalpha.com/input/?i=int%28s^3+%2F+%28s^2+%2B+a^2%29^3+%2C+s+%3D+0..infty%29" target="_blank" class="link link--external" rel="nofollow ugc noopener">here</a>. Incindentally, WolframAlpha doesn't know what to do with this integral: $\int_{a^2}^{\infty} \frac{u - a^2}{u^3} du$, (which occurs after my substitution) see <a href="http://www.wolframalpha.com/input/?i=int%28+%28u+-+a^2%29+%2F+u^2%2C+u+%3D+a^2..infty%29" target="_blank" class="link link--external" rel="nofollow ugc noopener">here</a>, which I find to be strange and amusing. What's going on here? I don't understand, and I feel rather silly that this straightforward integral is causing me problems.[/itex]

 

[CompuChip]

If $du = 2 s ds$ then
$$\frac{1}{2} \frac{s^2}{u^3} (2 s ds) = \frac{1}{2} \frac{u^2 - a^2}{u^3}$$
Note that this differs from your expression by a square on the u.

 

[JonF]

I believe you messed up your substitution in the numerator/differential. I personally would use the sub: s=atanθ. This will give you an integral that is easier to deal with, After that sub integration by parts twice should get you your answer.

 

[fluxions]

$$\int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds <br /> = \frac{1}{2}\int_{a^2}^{\infty} \frac{s^2}{u^3} du<br /> = \frac{1}{2}\int_{a^2}^{\infty} \frac{u - a^2}{u^3} du$$
right, since $$s^2 = u - a^2$$ by how I defined u.

 

[JonF]

Wow, sorry I wasn’t paying attention at all since I’m so used to squared terms in the denominator being a trig sub.

Your error is when you evaluate the integral. I get 1/2(-1/u + a^2/2u^2) evaluated at inf and a^2 which works out to your desired result.

 

[fluxions]

Wow, sorry I wasn’t paying attention at all since I’m so used to squared terms in the denominator being a trig sub.

Your error is when you evaluate the integral. I get 1/2(-1/u + a^2/2u^2) evaluated at inf and a^2 which works out to your desired result.

Yes, I made a typo. I have corrected it. There is still a problem, however.

 

[CompuChip]

Isn't that the same as what fluxions wrote (apart from a typo in the second-to-last identity, where there is u instead of 1/u, but which is corrected afterwards)?

Making such stupid mistakes, I refrain from replying more, and will get some sleep :)

 

[CompuChip]

Wolfram can do the integral however: http://www.wolframalpha.com/input/?i=integrate+(u+-+a^2)+/+u^3+from+u+=+a^2+to+infinity

 

[CompuChip]

I just spotted one of the mistakes...
$$\left( \frac{1}{u} - \frac{1}{2} \frac{a^2}{u^2} \right)_{u = a^2} = \frac{1}{a^2} - \frac{a^2}{2 (a^2)^2} = \frac{1}{a^2} - \frac{1}{2} \frac{1}{a^2} = \frac{1}{2} \frac{1}{a^2}$$

I completely overlooked that you square the square in the second term.

 

[fluxions]

Wolfram can do the integral however: http://www.wolframalpha.com/input/?i=integrate+(u+-+a^2)+/+u^3+from+u+=+a^2+to+infinity

Interesting. It gives a different answer that it does for the 'original' integral. And, when I tried to get Woflram to evaluate the 'substituted' integral I used the EXACT same syntax as I did for the 'original' integral...

 

[fluxions]

I just spotted one of the mistakes...
$$\left( \frac{1}{u} - \frac{1}{2} \frac{a^2}{u^2} \right)_{u = a^2} = \frac{1}{a^2} - \frac{a^2}{2 (a^2)^2} = \frac{1}{a^2} - \frac{1}{2} \frac{1}{a^2} = \frac{1}{2} \frac{1}{a^2}$$

I completely overlooked that you square the square in the second term.

Indeed. Problem solved. Thanks.