# Evaluating $\int_0^z(z-x)^{2m-1}x^{2m-1}e^{-\frac{m}{\Omega}x^2}\times K_0(\frac{2(z-x)}{\alpha}\frac{m}{\Omega}){\rm d}x$

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.

• Who gave you such a monstrosity to integrate? – Kenny Lau Mar 11 at 1:05
• Yeah, it's as monstrous as my Majin form... – Vegeta the Prince of Saiyans Mar 11 at 1:14
• @kenny I obtained it by solving this: A+BC where A,B,C are nakagami random variables.... – Pranu Mar 11 at 3:35
• @metamorphy, oh....my mistake it should be z and not infinity.... – Pranu Mar 11 at 4:51

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.
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 $$\infty$$