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I am trying to solve this integral which has come up as part of some other work, but it is proving to be much harder than I had originally thought. For $0 < |a| \le 1$ being some constant, I am trying to solve the following

$$f_Z(z)=\lim_{z\to \infty} \int_0^z{\dfrac{\exp{\left(-\left(\frac{z-w}{2} + \frac{w}{2a^2}\right)\right)}}{\sqrt{\left(z - w\right)w}}}dw$$

which I reduced to $$f_Z(z)=\lim_{z\to \infty} \exp{\left(\frac{-z}{2}\right)}\int_0^z{\dfrac{\exp{\left(\frac{w\left(a^2 - 1\right)}{2a^2}\right)}}{\sqrt{\left(z - w\right)w}}}dw$$

Using the substitution $t=\sqrt{\left(z - w\right)}$ which gives $w=z - t^2$ and $\frac{dw}{dt}=-2t$. I was able to restate it as $$f_Z(z)=\lim_{z\to \infty}\exp{\left(\frac{-z}{2}\right)}\int_\sqrt{z}^0{\dfrac{-2\exp{\left(\frac{\left(z - t^2\right)\left(a^2 - 1\right)}{2a^2}\right)}}{\sqrt{\left(z - t^2\right)}}}dt$$ Using the substitution $t=\sqrt{z}\sin \theta$ and $\frac{dt}{d\theta}=\sqrt{z}\cos \theta$ and the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$. I was able to restate it as. $$f_Z(z)=\lim_{z\to \infty}\exp{\left(\frac{-z}{2}\right)}\int_\frac{\pi}{2}^\pi{-2\exp{\left(z\cos^2 \theta \frac{\left(a^2 - 1\right)}{2a^2}\right)}}d\theta$$.

I am not quite sure how to proceed here. For all I know, trying to get rid of those square root parts, I may have made this worse, not only I am unsure about how to deal with the cosine in the power part of the $e$, I also have this negative sign to deal with now.

Any help would be much appreciated.

Edit: Using Laplace transform and then the inverse I was able to compute this to $$f_Z(z)=\dfrac{exp\left(\frac{-z}{4}\left(1 + \frac{1}{a^2}\right)\right)}{4a^2}J_0\left(\dfrac{z\sqrt{1-a}\sqrt{a+1}\sqrt{a^2 -1}}{2a}\right)$$ where $J_0$ is the Bessel function of first kind. The link for representation Wolfram Alpha Inverse Laplace Transform. I would love to see a derivation without the Laplace transform usage. :-)

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  • $\begingroup$ If $z$ and $a$ are constants, then for sake of simplicity the relevant integral is just $\int e^{\mu \cos^2\theta}d\theta$. This should be easier to find in a table (I"m almost positive it's a Bessel function identity.) $\endgroup$ – Semiclassical Jul 16 '14 at 20:21
  • $\begingroup$ Do you have particular bounds of integration in mind? $\endgroup$ – Semiclassical Jul 16 '14 at 20:34
  • $\begingroup$ Sorry I was driving so could not reply earlier the z is in fact a variable but obviously a constant as far as the integral in concerned. Both z and w I think can be treated as non negative reals. Could please cite me such a table on the web. $\endgroup$ – Comic Book Guy Jul 16 '14 at 22:21
  • $\begingroup$ Do you seek antiderivative? Or there are limits? Please specify! $\endgroup$ – Mohammad W. Alomari Jul 19 '14 at 12:21
  • $\begingroup$ @mwomath, thank you for your query, I actually went back to equation I was deriving and much earlier in my derivation there was a convolution integral step, instead of trying to compute that by substitutions, which got me this problem, I instead used laplace transforms and inverse laplace transforms to get the appropriate Bessel function based probability density I was trying to get. I 'll update the integral with the limits shortly. $\endgroup$ – Comic Book Guy Jul 19 '14 at 21:13
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For the integral $$I:=\int_0^z {\frac {\exp {(w\cdot \frac{a^2-1}{2a^2})}}{\sqrt {(z-w) w}} dw} $$ by completing the square in the denumenator and simplify we get $$\sqrt{\frac{1}{4} z-\frac{1}{4} z+zw-w^2}=\frac{1}{2} z\sqrt {1-(1-\frac{2w}{z})^2}$$ Now using the substitution $$\sin\theta = 1-\frac {2w}{z} $$ Therefore $ I $ reduces to $$ I=\int_{\frac {-\pi}{2}}^{\frac{\pi}{2}}{ e^{\alpha_z (1-\sin\theta)}d\theta}$$ where $\alpha_z=\frac{z}{2}\frac{a^2-1}{2a^2} $. Now, expand $\alpha_z (1-\sin\theta) $ in exponential Taylor series we get $$ I=\int_{\frac {-\pi}{2}}^{\frac {\pi}{2}}{\sum_n { \frac {\alpha_z^n}{n!}(1-\sin\theta)^n }d\theta}.$$ assuming the uniform convergence of the series then term by term integration yields that $$ I=\sum_n {\frac{\alpha_z^n}{n!}\cdot \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} {(1-\sin\theta)^n d\theta}} $$ Using binomial series we may write $$(1-\sin\theta)^n=\sum_{k=0}^{n}{(-1)^k \left ({\begin{array}{*{20} c} n \\ k \\ \end{array}}\right) \int_{\frac {-\pi}{2}}^{ \frac {\pi}{2 }}{\sin^{n-k}\theta d\theta}}.$$

It's well known that $\int\sin^m x$ given by:

If $ m$ even i.e., $ m =2n $ then $$ \int{\sin^{2n} xdx}= 2^{-2n} \left ({\begin {array}{*{20} c} 2n \\ n \\ \end {array}}\right) x+(-1)^{n} 2^{1-2n}\sum_{ k=0 }^{n-1}(-1)^k\left ({\begin {array}{*{20} c} 2n\\ k\\ \end{array}}\right) \frac {\sin (2n-2k) x}{2n-2k} $$ and for $ m $ odd; $ m=2n+1$ we have $$\int { \sin^{2n+1} xdx }=2^{-2n}(-1)^n \sum_{k=0}^n \left ({\begin {array}{*{20} c} 2n+1\\ k\end {array}}\right) \frac { \cos (2n+1-2k) x }{2n+1-2k} $$

For the both cases substitute the result in the original sum and simplify to get the result. It remains to calculate the limit as $ z→\infty $.

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