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How can I integrate the following:

$$\frac{1}{b^2}\int_0^\infty z^{-2}\exp(-a z)\sin^2(b z)\, \mathrm dz$$

for $a,b>0$? Maple gives a compact result:

$$\frac{1}{b} \tan^{-1}(c) - \frac{1}{ac^2} \ln(1 + c^2)$$

where $c=2b/a$. I can solve it when the powers of $z$ and $\sin$ are $-1$ and $1$ respectively, but not for 2.

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1 Answer 1

up vote 3 down vote accepted

Let $I(a)$ denote the integral. Consider $$\begin{eqnarray} I^{\prime\prime}(a) &=& \frac{1}{b^2} \int_0^\infty \exp(-a z) \sin^2(b z) \mathrm{d}z = \frac{1}{2b^2} \int_0^\infty \exp(-a z) (1-\cos(2 b z)) \mathrm{d}z \\ &=& \frac{1}{2b^2} \Re \int_0^\infty (\exp(-a z)-\exp(-(a+ 2i b) z)) \mathrm{d}z = \frac{1}{2b^2} \Re\left(\frac{1}{a} - \frac{1}{a+2 i b}\right) = \frac{2}{a} \frac{1}{a^2 + 4 b^2} \end{eqnarray} $$ Now integrate back, using $I(\infty) = 0$ and $I^\prime(\infty)=0$: $$ I^{\prime}(a) = \int_{a}^\infty \frac{2}{t} \frac{1}{t^2 + 4 b^2} \mathrm{d}t = \frac{1}{4 b^2} \log\left(1+ \frac{4b^2}{a^2}\right) $$ $$ I(a) = \int_{a}^\infty I^{\prime}(t) \mathrm{d}t = \frac{1}{b} \arctan\left(\frac{2b}{a}\right) - \frac{a}{4b^2} \log\left(1+ \frac{4b^2}{a^2}\right) $$

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Yes, it is differentiation with respect to a, twice. Interchange integration and differentiation with respect to parameter, and use $\frac{d^2}{d a^2} \exp(-a z) = z^2 \exp(-a z)$ –  Sasha Sep 17 '12 at 5:25

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