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I am trying to proceed with this integration but unable to solve completely $$f_{Z}(z)=\int_0^z(z-x)^{2m-1}x^{2m-1}e^{-\frac{m}{\Omega}x^2}\times K_0\left(\frac{2(z-x)}{\alpha}\frac{m}{\Omega}\right){\rm d}x$$ where $K_0$ is zeroth order bessel function.

Any help in this regard is highly appreciated.

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    $\begingroup$ Who gave you such a monstrosity to integrate? $\endgroup$ – Kenny Lau Mar 11 at 1:05
  • $\begingroup$ Yeah, it's as monstrous as my Majin form... $\endgroup$ – Vegeta the Prince of Saiyans Mar 11 at 1:14
  • $\begingroup$ @kenny I obtained it by solving this: A+BC where A,B,C are nakagami random variables.... $\endgroup$ – Pranu Mar 11 at 3:35
  • $\begingroup$ @metamorphy, oh....my mistake it should be z and not infinity.... $\endgroup$ – Pranu Mar 11 at 4:51
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so the zeroth order Besselfunction of the first kind is going with the series of ~$\sum_{m=0}^{\infty}(z-x)^{2m}$ and then some constant factors... i leave them out, u have to figure them out on ur own.

note that $\partial_x e^{x^2}$ equals $2x*e^{x^2}$ and the integral of $e^{-x^2}$ is the error function. Also note that $\partial_a e^{ax^2}|_{a=1}=x^2e^{x^2}$ so every x powered with a multiple of 2 can be replaced with a $\partial_a$ if u put an $a=1$ into the exponent and then it can be dragged out of the integral because $\partial_a$ and $\int dx$ commute. Dont forget to set a=1 after u are done.

I wont calculate this integral fully for u, but this could make it considerably easier. It is basically integration by differentiation. U just have to keep track of the constants that u get with every derivative.

consider also, that u have very simple integral limits... so when u have an uneven power of x as factor, u get a $e^{-ax^2}/a$ solution which is in this limits equal to $1/2a$ which u then just have to differentiate multiple times by $a$

one example: $\int dx (x^2+x^3)e^{-x^2}=\int dx (-\partial_a-\partial_a *x)e^{-ax^2}|_{a=1}=-\partial_a\int dx e^{-ax^2}-\partial_a \int dx$ $ x*e^{-ax^2}|_{a=1}=-\partial_a \sqrt\pi/(2\sqrt a)-\partial_a(1/2a)|_{a=1}=\sqrt \pi/4+1/2 $ in the limits of $0$ and $\infty$

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