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I'm having trouble with the substitutions in the following integral:

$$\int_0^\pi \int_0^\infty e^{-ibx\cos\theta - {x/a}} x^2 \sin\theta ~dx~d\theta $$

My attempt:

Let $$u = \cos\theta$$ then $$ du = -\sin\theta~d\theta$$

Then we have

$$-1 \int_0^\pi \int_0^\infty e^{-ibxu - {x/a}} x^2 ~du~dx $$

How do I separate the u out of the exponential to integrate separately with respect to $u$ and then $x$? Am I doing the wrong substitution?

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    $\begingroup$ Seems you forgot to change the bounds after the substitution. $\endgroup$ – Simply Beautiful Art Dec 6 '17 at 2:26
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    $\begingroup$ When you change variables, you have to change the endpoints to match: $\theta = 0$ corresponds to $u = \cos \theta = 1$, $\theta = \pi$ to $u = -1$. $\endgroup$ – Robert Israel Dec 6 '17 at 2:26
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    $\begingroup$ ... and then you can do the $du$ integral, treating $x$ as constant. $\endgroup$ – Robert Israel Dec 6 '17 at 2:31
  • $\begingroup$ Why do we treat x as constant? $\endgroup$ – Ella Dec 6 '17 at 2:33
  • $\begingroup$ @Ella As $u$ changes, $\theta$ changes, but does $x$ change? $\endgroup$ – Simply Beautiful Art Dec 6 '17 at 2:43
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\begin{align} \int_0^\pi \int_0^\infty e^{-ibx\cos\theta - \frac1ax} x^2 \sin\theta ~dx~d\theta &= \int_{-1}^1\int_0^\infty e^{-ibxu - \frac{x}{a}} x^2 ~dx ~du \\ &= \int_0^\infty\left(\int_{-1}^1 e^{-ibxu - \frac{x}{a}} x^2 ~du\right) ~du \\ &= \int_0^\infty x^2e^{-\frac{x}{a}}\left(\dfrac{e^{-ibx}}{ibx} - \dfrac{e^{-ibx}}{ibx}\right) ~du \\ &= \dfrac{2}{b}\int_0^\infty xe^{-\frac1ax}\sin bx dx \\ &= \dfrac{2}{b}{\cal L}(x\sin bx)\Big|_{s=\frac1a} \\ &= \dfrac{2}{b}\dfrac{2\frac1a~b}{(\frac{1}{a^2}+b^2)^2} \\ &= \dfrac{4a^3}{(1+a^2b^2)^2} \end{align}

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