MELLIN TRANSFORM OF MODIFIED BESSEL FUNCTION OF THE SECOND KIND

      Today we will proof the beautiful integral involving the modified Bessel function of the second kind, namely:


\int_{0}^{\infty}t^{s-1}K_{\nu}(t)dt=2^{s-2}\,\,\Gamma\left(\frac{s}{2}+\frac{\nu}{2} \right)\Gamma\left(\frac{s}{2}-\frac{\nu}{2} \right)


I=\int_{0}^{\infty}x^{s-1 }K_{\nu}(x)dx

We substitue the integral representation of  K_{\nu}(x)   proved here, then


I=\int_{0}^{\infty}x^{s-1 }\int_0^{\infty} e^{-x \cosh t} \cosh (\nu t) dt\,dx

=\int_{0}^{\infty}\cosh (\nu t) \int_0^{\infty} e^{-x \cosh t}x^{s-1 } dx\,dt

=\int_{0}^{\infty}\cosh (\nu t)\cdot \frac{\Gamma\left(s\right)}{\left( \cosh t\right)^{s}} dt

=\Gamma\left(s\right)\int_{0}^{\infty} \frac{\cosh (\nu t)}{ \cosh ^{s}(t)} dt

=\frac{\Gamma\left(s\right)}{2}\int_{0}^{\infty} \frac{e^{\nu t}+e^{-\nu t}}{ \cosh ^{s}(t)} dt

=2^{s-1}\Gamma\left(s\right)\int_{0}^{\infty} \frac{e^{\nu t}+e^{-\nu t}}{ \left( e^t+e^{-t} \right)^{s}} dt

=2^{s-1}\Gamma\left(s\right)\int_{0}^{\infty}\frac{e^{-st}}{e^{-st}}\cdot \frac{e^{\nu t}+e^{-\nu t}}{ \left( e^t+e^{-t} \right)^{s}} dt

=2^{s-1}\Gamma\left(s\right)\int_{0}^{\infty} \frac{e^{-(s-\nu) t}+e^{-(\nu+s) t}}{ \left( 1+e^{-2t} \right)^{s}} dt


Now, let     e^{-2t}=w\,\,\Rightarrow \,\, dt=-\frac{1}{2} \cdot\frac{dw}{w}


=2^{s-2}\Gamma\left(s\right)\int_{0}^1\frac{w^{\frac{(s-\nu)}{2}-1 }+w^{\frac{(\nu+s)}{2}-1 }}{ \left( 1+w \right)^{s}} \, dw


The last integral is a representation of the Beta function proved below in the appendix, then


I=2^{s-2}\Gamma\left(s\right)\frac{\Gamma\left(\frac{s}{2}+\frac{\nu}{2} \right)\Gamma\left(\frac{s}{2}-\frac{\nu}{2} \right)}{\Gamma\left(s \right)}


And we finally obtain


\boxed{\int_{0}^{\infty}t^{s-1}K_{\nu}(t)dt=2^{s-2}\,\,\Gamma\left(\frac{s}{2}+\frac{\nu}{2} \right)\Gamma\left(\frac{s}{2}-\frac{\nu}{2} \right)}(1)

Special case, let \nu=0 in (1) we obtain


\boxed{\int_{0}^{\infty}t^{s-1}K_{0}(t)dt=2^{s-2}\Gamma^2\left(\frac{s}{2} \right)}(2)


If we change s \mapsto 2s then make the change of variable t \mapsto 2x we get an integral representation for \Gamma^2\left(s \right)} in terms of Bessel functions:


\int_{0}^{\infty}t^{2s-1}K_{0}(t)dt=2^{2s-2}\Gamma^2\left(s \right)


2\int_{0}^{\infty}(2x)^{2s-1}K_{0}(2x)dx=2^{2s-2}\Gamma^2\left(s \right)


\boxed{4\int_{0}^{\infty}x^{2s-1}K_{0}(2x)dx=\Gamma^2\left(s \right)}(3)

Appendix

\int_0^{1}\frac{t^{x-1}+t^{y-1}}{\left(1+t \right)^{x+y}}dt=\frac{\Gamma\left(x \right)\Gamma\left(y \right)}{\Gamma\left(x+y \right)}

Proof:

Recall the integral representation for the Beta function

\int_0^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt=\frac{\Gamma\left(x \right)\Gamma\left(y \right)}{\Gamma\left(x+y \right)}

Then,

\int_0^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt=\int_0^{1}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt+\int_1^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt

let t=\frac{1}{w} in the second integral

=\int_0^{1}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt+\int_1^{0}\frac{w^{1-x}}{\left(1+\frac{1}{w} \right)^{x+y}}\frac{-dw}{w^2}

=\int_0^{1}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt+\int_0^{1}\frac{w^{1-x}w^{x+y}w^{-2}}{\left(1+w \right)^{x+y}}dw

\int_0^{\infty}\frac{t^{x-1}}{\left(1+t \right)^{x+y}}dt=\int_0^{1}\frac{t^{x-1}+t^{y-1}}{\left(1+t \right)^{x+y}}dt

Comments

Post a Comment

Popular posts from this blog