I am trying to understand the result of an integration obtained by using Mathematica software.
$p(x) = \int_0^{\gamma}\frac{2}{A\alpha^2g^2f^2}K_0\left(2\sqrt{\frac{x}{A\alpha^2g^2f^2}}\right)\text{d}x$-----(A)
where $K_0$ is zeroth order modified Bessel function of the second kind and $A,\alpha^2,g^2,f^2,\gamma$ are constant.
The Mathematica solved the equation (A) as
$P(\gamma) = \left[1-\frac{2\gamma K_1\left(2\sqrt{\frac{\gamma}{A\alpha^2g^2f^2}}\right)}{A\alpha^2g^2f^2\sqrt{\frac{\gamma}{A\alpha^2g^2f^2}}}\right]$----(B)
where $K_1$ indicates first order modified Bessel function of the second kind.
I am not getting how eq.(B) is obtained by Mathematica from eq.(A).
Any help in this regard is highly appreciated...
$$p(x) = \int_0^{\gamma}\frac{2}{A\alpha^2g^2f^2}K_0\left(2\sqrt{\frac{x}{A\alpha^2g^2f^2}}\right)\text{d}x$$
$$a=\frac 2 {\sqrt{A\alpha^2g^2f^2}}\implies p(x)=\frac 12 a^2\int K_0\left(a\sqrt{ x}\right)\,dx$$
Now $$a\sqrt{ x}=y\implies x=\frac{y^2}{a^2}\implies dx=\frac{2y}{a^2}\,dy\implies \int K_0\left(a\sqrt{ x}\right)\,dx=\frac 2{a^2} \int y K_0(y)\,dy$$ $$\int y K_0(y)\,dy=-y K_1(y)$$
Back to $x$ $$p(x)=-a \sqrt{x} K_1\left(a \sqrt{x}\right)$$ Using the bounds $$P(\gamma)=1-a \sqrt{\gamma } K_1\left(a \sqrt{\gamma }\right)$$ which seems to be what Mathematica gave (but you need to simplify its result).