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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.

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  • $\begingroup$ No, you can't. The first integral is basically indefinite. $\endgroup$
    – metamorphy
    Mar 13, 2021 at 10:36
  • $\begingroup$ Ok. Could you please tell the approach to solve this integration. $\endgroup$
    – Pranu
    Mar 13, 2021 at 10:49

1 Answer 1

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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.

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    $\begingroup$ Can you please tell me the procedure to solve this type of integration. $\endgroup$
    – Pranu
    Mar 13, 2021 at 10:54

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