Ok, so I need to evaluate this integral: [tex] \int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds [/tex](adsbygoogle = window.adsbygoogle || []).push({});

It seems evident that a substitution ought to do the trick. Let's try [itex] u = s^2 + a^2 [/itex], so [itex] du = 2sds [/itex]. Hence

[tex] \int_0^{\infty} \frac{s^3}{(s^2 + a^2)^3} ds

= \frac{1}{2} \int_{a^2}^{\infty} \frac{u - a^2}{u^3} du

= \frac{1}{2} \int_{a^2}^{\infty} \frac{1}{u^2} - \frac{a^2}{u^3} ds

= \frac{1}{2} \left(-\frac{1}{u} + \frac{1}{2} \frac{a^2}{u^2}\right)\right|_{a^2}^{\infty}

= \frac{1}{2} \left(\frac{1}{a^2} - \frac{1}{2}\right) [/tex]

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 here. Incindentally, WolframAlpha doesn't know what to do with this integral: [itex] \int_{a^2}^{\infty} \frac{u - a^2}{u^3} du [/itex], (which occurs after my substitution) see here, 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.

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# Straightforward integration

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