# How to integrate a modified Bessel function of the second kind times a polynomial?is it possible to have a closed-form expression?

what is the steps of evaluating this integral: $$\int_{0}^{x} x^{M-1}K_0(2\sqrt{x}) dx$$. where $$M$$ is a positive integer and $$K_0(.)$$ is the modified Bessel function of the second kind

• Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Commented Apr 23, 2023 at 9:30
• This is integral from $0$ to $\omega$? That is, an indefinite integral? Or perhaps you want from $0$ to $\infty$, a definite integral? Commented Apr 23, 2023 at 9:35
• With Maple: $$\int_0^x t^{m-1} K_0\left(2 \sqrt{t}\right) \, dt=\frac{1}{2} x G_{1,3}^{2,1}\left(x\left|\begin{array}{c} 0 \\m-1,m-1,-1 \\\end{array}\right.\right)$$ for $x\in \mathbb{R}$ and G is MeijerG function. Commented Apr 23, 2023 at 18:51

See that, $$\int_0^a x^{n-1}K_0(2\sqrt x)\mathrm dx \\ =\int_0^{2\sqrt a} \left(\frac{t^2}{4}\right)^{n-1}K_0(t) \frac{t}{2}\mathrm dt$$

So you can consider the equivalent problem $$\int_0^z x^{2n-1}K_0(x)\mathrm dx$$

In the case that $$z=+\infty$$ we have (see 10.43.19 ) $$\int_0^\infty x^{\mu-1}K_\nu(x)\mathrm dx= 2^{\mu-2}~\Gamma(\mu)\mathrm B\left(\frac{\mu+\nu}{2},\frac{\mu-\nu}{2}\right)\\ |\Re(\nu)|<\Re(\mu)$$ So, in the case that $$\nu=0~,~\mu=2n-1$$, we have $$\int_0^\infty x^{2n-1}K_0(x)\mathrm dx=2^{2n-2}\Gamma(2n)~\mathrm B\left(n,n\right) \\ =4^{n-1}\Gamma(2n)~\frac{\Gamma(n)^2}{\Gamma(2n)} \\ =4^{n-1}~(n-1)!^2$$ For general (large) $$z$$ you can write

$$\int_0^z x^{2n-1}K_0(x)\mathrm dx=\int_0^\infty x^{2n-1}K_0(x)\mathrm dx-\int_z^\infty x^{2n-1}K_0(x)\mathrm dx \\ \int_0^z x^{2n-1}K_0(x)\mathrm dx=4^{n-1} (n-1)!^2-\int_z^\infty x^{2n-1}K_0(x)\mathrm dx$$

To estimate the integral on the right, use the bounds (see Corollary 3.4 of this) $$\sqrt{\frac{\pi}{2}}\frac{\mathrm e^{-x}}{\sqrt{x+1/2}}0$$ To say that (using the cruder lower bound $$0 $$\int_z^\infty x^{2n-1}K_0(x)\mathrm dx < \int_z^\infty x^{2n-1/2}\sqrt{\frac{\pi}{2}}\mathrm e^{-x}\mathrm dx \\ \int_z^\infty x^{2n-1}K_0(x)\mathrm dx< \sqrt{\frac{\pi}{2}}\Gamma\left(2n+\frac{1}{2},z\right)$$ Where $$\Gamma(\cdot,\cdot)$$ is an upper incomplete Gamma function.

So you can establish the bounds $$\boxed{4^{n-1}(n-1)!^2 < \int_z^\infty x^{2n-1}K_0(x)\mathrm dx < 4^{n-1}(n-1)!^2-\sqrt{\frac{\pi}{2}}\Gamma\left(2n+\frac{1}{2},z\right)}$$

To improve this you will need to use the more precise lower bound $$\sqrt{\frac{\pi}{2}}\frac{\mathrm e^{-x}}{\sqrt{x+1/2}} but that results in a fairly intractable integral.