Triple integral of $r^{2}e^{ir\cos\theta}\sin\theta \,dr\,d\theta \,d\phi$

I'm trying to calculate this integral but I'm a bit stuck. Has anyone got any tips/tricks to deal with the $e^{ir\cosθ}$ part?

$$\iiint r^{2}e^{ir\cos\theta}\sin\theta \,dr\,d\theta \,d\phi$$

Limits: $0\leq r \leq a$, $0\leq\theta\leq \pi$, $0\leq\phi\leq2\pi$.

I'm a first year chemistry student so keep the maths as simple as possible!

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Your integral can be rewritten (by Fubini): $$\left(\int_{\phi=0}^{2\pi}d\phi\right)\left(\int_{r=0}^ar^2\left(\int_{\theta=0}^\pi e^{ir\cos\theta}\sin\theta d\theta\right)dr\right)$$ Of course, the first factor is $2\pi$. Now for every $r>0$, do the change of variable $u=ir\cos\theta$, $du=-ir\sin\theta d\theta$ in the middle integral to get $$\int_{\theta=0}^\pi e^{ir\cos\theta}\sin\theta d\theta=\int_{ir}^{-ir}e^u\frac{-du}{ir}=\frac{1}{ir}\int_{-ir}^{ir}e^udu=\frac{1}{ir} e^u\rvert_{-ir}^{ir}= \frac{1}{ir}(e^{ir}-e^{-ir}).$$ Now $$\int_{r=0}^ar^2\left(\int_{\theta=0}^\pi e^{ir\cos\theta}\sin\theta d\theta\right)dr=\frac{1}{i}\int_0^ar(e^{ir}-e^{-ir})dr=2\int_0^ar\sin r dr$$ by Euler's formula $e^{ir}-e^{-ir}=2i\sin r$.

It only remains to integrate by parts $$\int_0^ar\sin dr=(-r\cos r)\rvert_0^a+\int_0^a\cos r dr=-a\cos a+\sin r\rvert_0^a=-a\cos a+\sin a.$$ Finally, your integral is worth $$2\pi\cdot2(\sin a-a\cos a)=4\pi(\sin a -a\cos a).$$

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And you wonder why I put my $d$'s next to my integrals? – Ron Gordon Mar 14 '13 at 19:31
@RonGordon He he he...I see it now. – 1015 Mar 14 '13 at 19:33

Hint: $$(e^{\cos{t}})'=-\sin{t} \cdot e^{\cos{t}}$$ Integrate with respect to $\theta$ first, you should get $ir(e^{ir}-e^{-ir})$.

Use the formula $$sinr=\frac{e^{ir}-e^{-ir}}{2i}$$

You should be able to come up with the rest.

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