Integrate Modified Bessel Function

**Karl Harder** · August 27th 2009, 05:52 AM

I have been given a modified Bessel function $I_{n}(z)$ and it it defined by the integral

$I_{n}(z) = \frac{1}{\pi}\int_{0}^{\pi}d\theta{e^{zCos\theta}} Cos(n\theta)$

I have been asked to show that

(a) $I_{n-1}(z) - I_{n+1}(z)=\frac{2n}{z}I_{n}(z)$

(b) $I_{n}(z)=I'_{N+1}(z)+\frac{n}{z}I_{n}(z)$

Firstly i have started by trying to integrate the integral. Using $u=e^{zCos\theta}$ and $dv=Cos(n\theta)d\theta$ . Therefore $du=-Sin\theta{e^{zCos\theta}}$ and $v=\int_{}cosn\theta{d\theta}=\frac{1}{n}Sin\theta{ n\theta}$

Using integration by parts $\int_{}udv=uv-\int_{}vdu$

I ended up with

$I_{n}(z) = \frac{1}{\pi}\int_{0}^{\pi}d\theta{e^{zCos\theta}} Cos(n\theta)=\frac{1}{n}Sin\theta{n\theta}e^{zCos\ theta} + \int_{}\frac{1}{n}Sin\theta{Sin\theta}{n\theta}{e^ {zCos\theta}}$

knowing that $sin\theta{sin\theta} = cos(n\theta+\theta) + cos(n\theta-\theta)$

Therefore

$I_{n}(z) = \frac{1}{n}Sin\theta{n\theta}e^{zCos\theta} + \int_{}\frac{1}{n}{cos(n\theta+\theta) + cos(n\theta-\theta)}{n\theta}{e^{zCos\theta}}$

I'm not sure if i'm on the right track to solving this problem. Any ideas of what i can do next? How do i then show that

(a) $I_{n-1}(z) - I_{n+1}(z)=\frac{2n}{z}I_{n}(z)$

(b) $I_{n}(z)=I'_{N+1}(z)+\frac{n}{z}I_{n}(z)$

**halbard** · August 27th 2009, 12:25 PM

Start with $I_{n-1}(z)-I_{n+1}(z)=\frac1\pi\int_0^\pi\mathrm e^{z\cos\theta}[\cos(n-1)\theta-\cos(n+1)\theta]\mathrm d\theta$ $=\frac2\pi\int_0^\pi\mathrm e^{z\cos\theta}\sin n\theta\,\sin\theta\,\mathrm d\theta=-\frac2{\pi z}\int_0^\pi\mathrm \sin\,n\theta\,\mathrm d(e^{z\cos\theta})$ .

Integrate the last term by parts, so that

$I_{n-1}(z)-I_{n+1}(z)=\left[-\frac2{\pi z}\sin n\theta\,e^{z\cos\theta}\right]_0^\pi+\frac{2n}{\pi z}\int_0^\pi e^{z\cos\theta}\cos n\theta\,\mathrm d\theta=\frac{2n}z I_n(z)$ .

For the second one, start with $I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos(n+1)\theta\,\cos\theta\,\mathr m d\theta$ .

Thus $I_n(z)-I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}[\cos n\theta-\cos(n+1)\theta\,\cos\theta]\mathrm d\theta$ $=\frac1\pi\int_0^\pi e^{z\cos\theta}\sin(n+1)\theta\,\sin\theta\,\mathr m d\theta=\frac{n+1}z I_{n+1}(z)$ (similar to the last part of the previous derivation).

Warning: the last identity is not quite the one given.

**Karl Harder** · August 29th 2009, 05:30 AM

What does it mean by I' (I prime) i can't seem to find much information about it.

**Danny** · August 29th 2009, 06:17 AM

Originally Posted by Karl Harder

What does it mean by I' (I prime) i can't seem to find much information about it.

It's the derivative of $I$ . If $I_{n+1}(z) = \frac{1}{\pi}\int_0^{\pi} e^{z \cos \theta} \cos (n+1) \theta\, d \theta$ then $I'_{n+1}(z) = \frac{1}{\pi}\int_0^{\pi} e^{z \cos \theta} \cos \theta \cos (n+1) \theta\, d \theta$ .

**Karl Harder** · August 29th 2009, 10:41 PM

Originally Posted by halbard

Start with $I_{n-1}(z)-I_{n+1}(z)=\frac1\pi\int_0^\pi\mathrm e^{z\cos\theta}[\cos(n-1)\theta-\cos(n+1)\theta]\mathrm d\theta$ $=\frac2\pi\int_0^\pi\mathrm e^{z\cos\theta}\sin n\theta\,\sin\theta\,\mathrm d\theta=-\frac2{\pi z}\int_0^\pi\mathrm \sin\,n\theta\,\mathrm d(e^{z\cos\theta})$ .

Integrate the last term by parts, so that

$I_{n-1}(z)-I_{n+1}(z)=\left[-\frac2{\pi z}\sin n\theta\,e^{z\cos\theta}\right]_0^\pi+\frac{2n}{\pi z}\int_0^\pi e^{z\cos\theta}\cos n\theta\,\mathrm d\theta=\frac{2n}z I_n(z)$ .

For the second one, start with $I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos(n+1)\theta\,\cos\theta\,\mathr m d\theta$ .

Thus $I_n(z)-I'_{n+1}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}[\cos n\theta-\cos(n+1)\theta\,\cos\theta]\mathrm d\theta$ $=\frac1\pi\int_0^\pi e^{z\cos\theta}\sin(n+1)\theta\,\sin\theta\,\mathr m d\theta=\frac{n+1}z I_{n+1}(z)$ (similar to the last part of the previous derivation).

Warning: the last identity is not quite the one given.

Does that mean that if

$I_{n}(z)=\frac1\pi\int_0^\pi e^{zcos\theta} \cos n\theta\, d\theta$

then

$I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\,\cos\theta\, d\theta$

**Danny** · August 30th 2009, 05:07 AM

Originally Posted by Karl Harder

Does that mean that if

$I_{n}(z)=\frac1\pi\int_0^\pi e^{zcos\theta} \cos n\theta\, d\theta$

then

$I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\,\cos\theta\, d\theta$

Yep, it sure does.

**Karl Harder** · August 30th 2009, 05:38 AM

I seem to be missing the $\cos{\theta}$ from the result. I'm not sure where i might have gone wrong. I have checked it 3 times each different ways and i still keep getting the same answer.

$I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\, d\theta$

I am using $I'_{n}(z) =I_{n+1}+\frac{n}{z} I_{n}(z)$

which can be rewritten as $I'_{n}(z) =I_{n+1}(z)+\frac{1}{2} I_{n-1}(z)-\frac{1}{2}I_{n+1}(z)$ which was prooven in the first part of this question.

Thus $I'_{n}(z) =\frac{1}{2}(I_{n+1}(z)+I_{n-1}(z))$

And then i put the values in for $I_{n+1}(z)$ and $I_{n-1}(z))$

and somehow i end up with the answer $I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\, d\theta$

Although it should be $I'_{n}(z)=\frac1\pi\int_0^\pi e^{z\cos\theta}\cos n\theta\,$ $\cos \theta\ d\theta$

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