# How to solve this type of integration?

I had the following integration: $$\int_0^z w^{m-1}K_0(2\frac{m}{\Omega}\sqrt{w})dw.\tag{1}$$

Can i use the following formula from Table of integrals to solve the integral in eq. $$(1)$$

$$\int_0^{\infty}x^{\mu}K_v(ax)dx=2^{\mu-1}a^{-\mu-1}\Gamma\left(\frac{1+\mu+v}{2}\right)\Gamma\left(\frac{1+\mu-v}{2}\right).$$ Any help in this regard is highly appreciated.

• No, you can't. The first integral is basically indefinite. – metamorphy Mar 13 at 10:36
• Ok. Could you please tell the approach to solve this integration. – Pranu Mar 13 at 10:49

## 1 Answer

It is much more complicated for finite values of $$z$$ since it will be (assuming $$m>0$$).

Let $$w=\frac{ \Omega ^2}{4 m^2}t^2$$ to face $$I=2 \left(\frac{\Omega ^2}{4m^2}\right)^m\int t^{2 m-1} K_0(t)\,dt$$ $$\int t^{2 m-1} K_0(t)\,dt=$$ $$\frac{t^{2 m} \left(2 m K_0(t) \, _1F_2\left(1;m,m+1;\frac{t^2}{4}\right)+t K_1(t) \, _1F_2\left(1;m+1,m+1;\frac{t^2}{4}\right)\right)}{4 m^2}$$

For $$t=0$$ the value is zero.

• Can you please tell me the procedure to solve this type of integration. – Pranu Mar 13 at 10:54