INTEGRAL REPRESENTATION FOR THE MODIFIED BESSEL FUNCTION OF THE SECOND KIND AND BASSET´S INTEGRAL

Today´s post is a very special one for me. It took me a very long time to proof these Bessel functions integral representations and be able to present them the way I show here . To begin, we start proving Hankel´s contour integral representation for the Gamma Function which serves as the basis to derive the other integral representations for the Bessel function. Then, we proof some integral representations for the Modified Bessel function of the first and second kind. Finally, we conclude the post computing Basset´s integral that shows up in probability as the characteristic function of the Student´s T distribution with the help of the integrals proved in the previous sections.

Hankel´s contour integral for The Gamma function

The Gamma function has the following integral representation:

$\frac{1}{\Gamma\left(z \right)}=\frac{1}{2 \pi i}\int_C e^{t}t^{-z}dt$ (1)

Proof: Let

$I=\int_C e^{t}t^{-z}dt=\int_{-\infty}^{-\epsilon}+\int_{-\pi}^{\pi}+\int_{-\epsilon}^{-\infty}\left(e^{t}t^{-z}dt\right)$

$\int_C e^{t}t^{-z}dt=I_1+I_2+I_3$ (2)

And be the following contour

For the first integral on the line below the negative Real axis we have

$I_1=\int_{-\infty}^{-\epsilon}e^{t}t^{-z}dt$

$\begin{aligned} &\text{let}\,\,t=Re^{-i \pi} \\ &dt=e^{-i \pi}dR=-dR\\ &\text{when}\,\,t=-\infty \, \Rightarrow R=\infty \\ &\text{when}\,\,t=-\epsilon \, \Rightarrow R=\epsilon \\ \end{aligned}$

Then

$I_1=\int_{-\infty}^{-\epsilon}e^{t}t^{-z}dt=\int_{\infty}^{\epsilon}\left( Re^{-i \pi}\right)^{-z}e^{Re^{-i \pi}}\left( -dR\right)$

$=e^{i z \pi}\int_{\epsilon}^{\infty}R^{-z}e^{-R}dR$ (3)

For the integral around the circle we have

$I_2=\int_{-\pi}^{\pi}e^{t}t^{-z}dt$

$\begin{aligned} &\text{let}\,\,t=\epsilon e^{i \theta} \\ &dt=\epsilon i e^{i \theta} d \theta\\ \end{aligned}$

$I_2=\int_{-\pi}^{\pi}e^{t}t^{-z}dt=i \epsilon \int_{-\pi}^{\pi}\left(\epsilon e^{i \theta} \right)^{-z}e^{\epsilon e^{i \theta}}e^{i \theta}d \theta$

$=i \epsilon^{1-z} \int_{-\pi}^{\pi} e^{-i z \theta} e^{i \theta}e^{\epsilon \left(\cos \theta + i \sin \theta \right)} d \theta$

$=i \epsilon^{1-z} \int_{-\pi}^{\pi} e^{-i (1-z) \theta} e^{\epsilon \left(\cos \theta + i \sin \theta \right)} d \theta$

$|I_2| \leq|\epsilon^{1-z}|\int_{-\pi}^{\pi}|e^{\epsilon \cos \theta}| d \theta$

taking the limit $\epsilon \longrightarrow 0$

$\lim_{\epsilon \longrightarrow 0} |I_2| \leq \lim_{\epsilon \longrightarrow 0} |\epsilon^{1-z}|\int_{-\pi}^{\pi}|e^{\epsilon \cos \theta}| d \theta$

$\lim_{\epsilon \longrightarrow 0} |I_2| \leq \lim_{\epsilon \longrightarrow 0} |\epsilon^{1-z}|\int_{-\pi}^{\pi}\lim_{\epsilon \longrightarrow 0}|e^{\epsilon \cos \theta}| d \theta$

$=\lim_{\epsilon \longrightarrow 0} |\epsilon^{1-z}|\int_{-\pi}^{\pi} d \theta$

$=2 \pi \lim_{\epsilon \longrightarrow 0} |\epsilon^{1-z}| \longrightarrow 0$

For the integral on the line above the negative real axis we have

$I_3=\int_{-\epsilon}^{-\infty}e^{t}t^{-z}dt$

$\begin{aligned} &\text{let}\,\,t=R e^{i \pi} \\ &dt= e^{i \pi} d R=-dR\\ \end{aligned}$

Then

$I_3=\int_{-\epsilon}^{-\infty}e^{t}t^{-z}dt=\int_{\epsilon}^{\infty}\left( Re^{i \pi}\right)^{-z}e^{Re^{i \pi}}\left( -dR\right)$

$=-e^{- i \pi z}\int_{\epsilon}^{\infty}e^{-R} R^{-z}dR$ (4)

Plugging (3) and (4) in (2) and already taking the limit $\epsilon \longrightarrow 0$ we obtain

$I=e^{i z \pi}\int_{0}^{\infty}R^{-z}e^{-R}dR -e^{- i \pi z}\int_{0}^{\infty}e^{-R} R^{-z}dR$

$I=\left(e^{i z \pi} -e^{- i \pi z}\right) \int_{0}^{\infty}e^{-R} R^{-z}dR$

$I=2 i \sin( z \pi)\Gamma \left(1-z \right)$

From the reflection formula

$\Gamma \left(z \right) \Gamma \left(1-z \right) =\frac{\pi}{\sin z \pi}$

We obtain

$I=2 i \sin( z \pi)\Gamma \left(1-z \right)=\frac{2 \pi i}{\Gamma \left(z \right) }$

and finally

$\boxed{\frac{1}{\Gamma\left(z \right)}=\frac{1}{2 \pi i}\int_C e^{t}t^{-z}dt}$ (5)

Contour Integrals representations for Modified Bessel Function

Now, recall the modified Bessel function:

$I_{\nu}(z)=\sum_{k=0}^{\infty} \frac{1}{k ! \Gamma(k+\nu+1)}\left(\frac{z}{2}\right)^{\nu+2 k}$ (6)

The function (6) can be defined or it may be also derived as a solution of a second order differential equation, here we take it as a definition. From (5) we can write

$\frac{1}{\Gamma(\nu+k+1)}=\frac{1}{2 \pi i}\int_{H} e^{t} t^{-(\nu+k+1)} d t$
(7)

H is the same contour as above. Substituting (7) in (6) we get

$I_{\nu}(z)=\frac{1}{2 \pi i} \sum_{k=0}^{\infty} \frac{1}{k !}\left(\frac{z}{2}\right)^{2 k}\left(\frac{z}{2}\right)^{\nu} \int_{H} e^{t} t^{-(\nu+1)} t^{-k} d t$

$I_{\nu}(z)=\frac{1}{2 \pi i}\left(\frac{z}{2}\right)^{\nu} \int_{H} e^{t} t^{-(\nu+1)} \sum_{k=0}^{\infty} \frac{t^{-k}}{k !} \cdot\left(\frac{z^{2}}{4}\right)^{k} d t$

$I_{\nu}(z)=\frac{1}{2 \pi i}\left(\frac{z}{2}\right)^{\nu} \int_{H} e^{t} t^{-(\nu+1)} e^{\frac{z^{2}}{4 t}} d t$

$\boxed{ I_{V}(z)=\frac{1}{2 \pi i}\left(\frac{z}{2}\right)^{v} \int_{H} t^{-(\nu+1)} e^{t+\frac{z^{2}}{4 t}} d t}}$ (9)

Now consider the following substitution in (9)

$t=\frac{z u}{2} \Rightarrow d t=\frac{z}{2} d u}$

and

$t+\frac{z^{2}}{4 t} \Rightarrow \frac{z}{2} u+\frac{z^{2}}{\frac{4}{2} z u}=\frac{z}{2} u+\frac{z}{2} \cdot u^{-1}=\frac{z}{2}\left(u+u^{-1}\right)$

Then

$I_{\nu}(z)=\frac{1}{2 \pi i}\left(\frac{z}{2}\right)^{\nu} \int_{H}\left(\frac{z}{2} \cdot u\right)^{-(v+1)} e^{\frac{z}{2}\left(u+u^{-1}\right)} \cdot \frac{z}{2} \cdot d u$

$I_{\nu}(z)=\frac{1}{2 \pi i}\left(\frac{z}{2}\right)^{\nu+1}\left(\frac{z}{2}\right)^{-(\nu+1)} \int_{H} u^{-(\nu+1)} e^{\frac{z}{2}\left(u+u^{-1}\right)} d u$

$\boxed{I_{\nu}(z)=\frac{1}{2 \pi i} \int_{H} u^{-(\nu+1)} e^{\frac{z}{2}\left(u+u^{-1}\right)} d u}$ (10)

Now we deform the contour H so that the radius of the circumference around the origin is , then (10) is rewritten as

$I_{\nu}(z)=\frac{1}{2 \pi i}\left\{\int_{-\infty}^{-1}+\int_{-\pi}^{\pi}+\int_{-1}^{-\infty}\left( u^{-(\nu+1)} e^{\frac{z}{2}\left(u+u^{-1}\right)} d u\right)\right\}=\frac{1}{2 \pi i}\left\{I_{2}+I_{2}+I_{3}\right\}$ (11)

$I_{1}: u=t e^{-i \pi} \Rightarrow d u=e^{-i \pi} d t$

$\begin{array}{l} u=-\infty \Rightarrow t=\infty \\ v=-1 \Rightarrow t=1 \end{array}$

$I_{1}=\int_{\infty}^{1}\left(t e^{-i \pi}\right)^{-(\nu+1)} e^{\frac{z}{2}\left(t e^{-i \pi}+t^{-1} e^{i \pi}\right)} e^{-i \pi} d t$

$I_{1}=-e^{i \pi \nu+i \pi-i \pi} \int_{1}^{\infty} t^{-(\nu+1)} e^{\frac{z}{2}\left(-t-t^{-1}\right)} d t$

$\boxed{I_{1}=-e^{i \pi v} \int_{1}^{\infty} t^{-(\nu+1)} e^{-\frac{z}{2}\left(t+t^{-1}\right)} d t}$ (12)

$I_{2}: u=e^{i \theta} \qquad d u=i e^{i \theta} d \theta$

$I_{2}=i \int_{-\pi}^{\pi}\left(e^{i \theta}\right)^{-(\nu +1)} e^{\frac{z}{2}\left(e^{i \theta}+e^{-i \theta}\right)} e^{i \theta} d \theta$

$I_{2}=i \int_{-\pi}^{\pi} e^{-i \nu \theta} e^{z \cos (\theta)} d \theta$

$\boxed{I_{2}=i \int_{-\pi}^{\pi} e^{z \cos (\theta)-i \nu \theta} d \theta}$ (13)

$I_{3}: u=te^{i \pi} \Rightarrow d u= e^{i \pi} d t$

$I_{3}=\int_{1}^{\infty} \left(te^{i \pi}\right)^{-(\nu+1)} e^{\frac{z}{2}\left(te^{i \pi}+t^{-1}e^{-i \pi}\right)}e^{i \pi} d t$

$\boxed{I_{3}=e^{-i \nu \pi}\int_{1}^{\infty} t^{-(\nu+1)} e^{-\frac{z}{2}\left(t+t^{-1}\right)} d t}$ (14)

Plugging (12), (13) and (14) in (11) we get

$I_{\nu}(z)=\frac{1}{2 \pi i}\left\{\left(e^{-i \pi \nu}- e^{i \pi \nu}\right) \int_{1}^{\infty} t^{-(v+1)} e^{-\frac{z}{2}\left(t+t^{-1}\right)} d t+i \int_{-\pi}^{\pi} e^{z \cos \theta-i v \theta} d \theta\right\}$

$\boxed{I_{\nu}(z)=\frac{-\sin (\nu \pi)}{\pi} \int_{1}^{\infty} t^{-(\nu+1)} e^{-\frac{z}{2}\left(t+t^{-1}\right)} d t+\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{z \cos \theta-i \nu \theta} d \theta}$ (15)

from (15) we can find $I_{-\nu}(z)$

$I_{-\nu}(z)=\frac{-\sin (- \nu \pi)}{\pi} \int_{1}^{\infty} t^{\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{z \cos \theta+i \nu \theta} d \theta$

$I_{-\nu}(z)=\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{z \cos \theta}\left(\cos (\nu \theta)+ i \sin (\nu \theta) \right) d \theta$

The first term becomes positive because of the oddness of sine

$I_{-\nu}(z)=\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{1}{2 \pi} \int_{-\pi}^{0} e^{z \cos \theta}\left(\cos (\nu \theta)+ i \sin (\nu \theta) \right) d \theta+\frac{1}{2 \pi} \int_{0}^{\pi} e^{z \cos \theta}\left(\cos (\nu \theta)+ i \sin (\nu \theta) \right) d \theta$

now let $\theta =-\phi$ in the first integral on the right hand side, then

$I_{-\nu}(z)=\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{1}{2 \pi} \int_{\pi}^{0} e^{z \cos -\phi}\left(\cos (-\nu \phi)+ i \sin (-\nu \phi) \right) (-d \phi)+\frac{1}{2 \pi} \int_{0}^{\pi} e^{z \cos \theta}\left(\cos (\nu \theta)+ i \sin (\nu \theta) \right) d \theta$

Since $\cos -\phi=\cos \phi \, \,\,\text{and}\,\,\, \sin - \phi=- \sin \phi \,\,\,\text {we get}$

$\boxed{I_{-\nu}(z)=\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{1}{ \pi} \int_{0}^{\pi} e^{z \cos \theta}\left(\cos (\nu \theta)\right) d \theta}$ (16)

Similarly we can write (15) as

$\boxed{I_{\nu}(z)=\frac{-\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{-\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{1}{ \pi} \int_{0}^{\pi} e^{z \cos \theta}\left(\cos (\nu \theta)\right) d \theta}$ (17)

Recall now the definition of the Modified Bessel function of the second kind

$K_{\nu}(z)=\frac{\pi}{2}\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin \nu \pi}$ (18)

So, if we subtract (11) from (10) we obtain

$I_{-\nu}(z)-I_{\nu}(z)=\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}+\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} t^{-\nu} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \frac{d t}{t}$

$I_{-\nu}(z)-I_{\nu}(z)=\frac{\sin ( \nu \pi)}{\pi} \int_{1}^{\infty} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \left(t^{\nu}+t^{-\nu} \right)\frac{d t}{t}$

$\frac{\pi}{2}\frac{I_{-\nu}(z)-I_{\nu}(z)}{\sin \nu \pi}=\frac{1}{2}\int_{1}^{\infty} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \left(t^{\nu}+t^{-\nu} \right)\frac{d t}{t}$

From (18) we conclude

$\boxed{K_{\nu}(z)=\frac{1}{2}\int_{1}^{\infty} e^{-\frac{z}{2}\left(t+t^{-1}\right)} \left(t^{\nu}+t^{-\nu} \right)\frac{d t}{t}}$ (19)

If we now let $t=e^{w} \Rightarrow dt=e^w dw$

$\begin{array}{l} t=1 \Rightarrow w=0 \\ t=\infty \Rightarrow w= \infty \end{array}$

$K_{\nu}(z)=\int_{0}^{\infty} e^{-\frac{z}{2}\left(e^w+e^{-w}\right)} \frac{\left(e^{\nu w}+e^{-\nu w} \right)}{2}\frac{e^w d w}{e^w}$

and Finally

$\boxed{K_{\nu}(z)=\int_{0}^{\infty} e^{-z \cosh w} \cosh (\nu w) dw}$ (20)

This result is shown here

Computing Basset’s Integral

Now, we are equipped to compute Basset´s integral given as following:

$I(kz)=\int_{0}^{\infty} \frac{\cos(kx)}{{\left(z^2+x^{2}\right)}^{\nu+\frac{1}{2}}}dx$ (20)

$I(kz)=\frac{1}{z^{2 \nu+1}}\int_{0}^{\infty} \frac{\cos(kx)}{{\left(1+\left(\frac{x}{z}\right)^{2}\right)}^{\nu+\frac{1}{2}}}dx$

Enforcing $x \longmapsto zx$ we obtain

$\int_{0}^{\infty} \frac{\cos(kx)}{{\left(z^2+x^{2}\right)}^{\nu+\frac{1}{2}}}dx=\frac{1}{z^{2\nu}}\int_{0}^{\infty} \frac{\cos(kzx)}{{\left(1+x^{2}\right)}^{\nu+\frac{1}{2}}}dx$ (21)

Lets for now consider that $kz=\lambda$ and just focus on the integral and call it , thus

$J=\int_{0}^{\infty} \frac{\cos(\lambda x)}{{\left(1+x^{2}\right)}^{\nu+\frac{1}{2}}}dx$

Now, consider the following integral, easily verified by a change of variable

$\Gamma\left(\nu+\frac{1}{2}\right)={\left(1+x^{2}\right)}^{\nu+\frac{1}{2}}\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{z-1}du$

now multiply by $\Gamma\left(\nu+\frac{1}{2}\right)$

$\Gamma\left(\nu+\frac{1}{2}\right) \cdot J=\int_{0}^{\infty}cos(\lambda x)\int_{0}^{\infty}e^{-\left(1+x^{2}\right)u} u^{\nu+\frac{1}{2}-1}\,du\,dx$

Swapping the integrals and distributing the exponential

$J=\frac{1}{\Gamma\left(\nu+\frac{1}{2}\right)}\int_{0}^{\infty}u^{\nu+\frac{1}{2}-1}e^{-u}\int_{0}^{\infty}e^{-x^{2}u}\cos( \lambda x)\,dx\,du$

The inner integral has the following solution proved previously here (equation 10)

$\int_{0}^{\infty}e^{-x^{2}u}\cos( \lambda x)\,dx=\frac{1}{2}\sqrt{\frac{\pi}{u}}e^{-\frac{\lambda^{2}}{4u}}$

Giving us:

$J=\frac{1}{\Gamma\left(\nu+\frac{1}{2}\right)}\frac{\sqrt{\pi}}{2}\int_{0}^{\infty}u^{\nu-1}e^{-u-\frac{\lambda^{2}}{4u}}\,\,du$

Now, lets make the following substitution $u=\left(\frac{\lambda}{2}\right)e^{w}$

$J=\frac{\sqrt{\pi}}{2}\frac{1}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{\lambda}{2}\right)}^{\nu}\int_{-\infty}^{\infty}e^{-\lambda\cosh(w)}e^{w\nu} dw$

$J=\frac{\sqrt{\pi}}{2}\frac{1}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{\lambda}{2}\right)}^{\nu}\left(\int_{-\infty}^{0}e^{-\lambda\cosh(w)}e^{w\nu} dw+\int_{0}^{\infty}e^{-\lambda\cosh(w)}e^{w\nu} dw\right)$

Let w=-t in the first integral of the R.H.S., then

$J=\frac{\sqrt{\pi}}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{\lambda}{2}\right)}^{\nu}\frac{1}{2}\left(\int_{\infty}^{0}e^{-\lambda\cosh(-t)}e^{-t\nu} (-dt)+\int_{0}^{\infty}e^{-\lambda\cosh(w)}e^{w\nu} dw\right)$

Since $\cosh (-w)=\cosh(w)$ we get the final result:

$\boxed{J=\frac{\sqrt{\pi}}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{\lambda}{2}\right)}^{\nu}\int_{0}^{\infty}e^{-\lambda\cosh(w)}\cosh{\left(w \nu\right)} dw}$ (22)

Now plugging (22) in (21) and recalling that $\lambda=kz$ we get

$\int_{0}^{\infty} \frac{\cos(kx)}{{\left(z^2+x^{2}\right)}^{\nu+\frac{1}{2}}}dx=\frac{1}{z^{2\nu}}\frac{\sqrt{\pi}}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{kz}{2}\right)}^{\nu}\int_{0}^{\infty}e^{-kz\cosh(w)}\cosh{\left(w \nu\right)} dw$

$\int_{0}^{\infty} \frac{\cos(kx)}{{\left(z^2+x^{2}\right)}^{\nu+\frac{1}{2}}}dx=\frac{\sqrt{\pi}}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{k}{2z}\right)}^{\nu}\int_{0}^{\infty}e^{-kz\cosh(w)}\cosh{\left(w \nu\right)} dw$

From (20) we get the beautiful result:

$\boxed{\int_{0}^{\infty} \frac{\cos(kx)}{{\left(z^2+x^{2}\right)}^{\nu+\frac{1}{2}}}dx=\frac{\sqrt{\pi}}{\Gamma\left(\nu+\frac{1}{2}\right)}{\left(\frac{k}{2z}\right)}^{\nu}K_{\nu}(kz) }$ (23)

Which matches with the result.

Special functions and their applications [by] N.N. Lebedev. Rev. English ed., translated and edited by Richard A. Silverman

G. N. Watson. A treatise on the Theory of Bessel Functions. Cambridge University Press, 1966.

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