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From: Han de Bruijn on 23 Jun 2010 08:53
integral(t = 0 .. 2*Pi) exp( A.cos(t) + B.sin(t) ) dt = ??

Han de Bruijn
From: dvsarwate on 23 Jun 2010 09:24
On Jun 23, 7:53 am, Han de Bruijn <umum...(a)gmail.com> wrote:
> integral(t = 0 .. 2*Pi) exp( A.cos(t) + B.sin(t) ) dt = ??
>
> Han de Bruijn

Since the argument of the exponential can be expressed
as sqrt(A^2 + B^2).cos(t - arctan(B/A)), the value should
be proportional to I_0(sqrt(A^2 + B^2)) where I_0 is the
0-th order modified Bessel function of the first kind.

http://www.math.sfu.ca/~cbm/aands/page_376.htm

Dilip Sarwate
From: Dan Cass on 23 Jun 2010 05:23
> integral(t = 0 .. 2*Pi) exp( A.cos(t) + B.sin(t) ) dt
> = ??
>
> Han de Bruijn

Maple doesn't even know the indefinite
integral of exp(cos(x)).

Nor does it know the definite integral
of exp(cos(x)) from 0 to 2*Pi.

This may mean that even a complex analysis contour integral approach will not work.
From: Han de Bruijn on 23 Jun 2010 10:24
On Jun 23, 3:23 pm, Dan Cass <dc...(a)sjfc.edu> wrote:
> > integral(t = 0 .. 2*Pi) exp( A.cos(t) + B.sin(t) ) dt
> > = ??
>
> > Han de Bruijn
>
> Maple doesn't even know the indefinite
> integral of exp(cos(x)).
>
> Nor does it know the definite integral
> of exp(cos(x)) from 0 to 2*Pi.
>
> This may mean that even a complex analysis contour integral approach will not work.

Well, I've seen some complex analysis in this group, just done by hand
(David Ullrich / World Wide Wade) where Maple couldn't keep up with ..

Han de Bruijn
From: Rob Johnson on 23 Jun 2010 11:47
In article <c641a0f7-31f3-4fe7-b393-b19339c2cb0e(a)5g2000yqz.googlegroups.com>,
dvsarwate <dvsarwate(a)gmail.com> wrote:
>On Jun 23, 7:53 am, Han de Bruijn <umum...(a)gmail.com> wrote:
>> integral(t = 0 .. 2*Pi) exp( A.cos(t) + B.sin(t) ) dt = ??
>>
>> Han de Bruijn
>
>Since the argument of the exponential can be expressed
>as sqrt(A^2 + B^2).cos(t - arctan(B/A)), the value should
>be proportional to I_0(sqrt(A^2 + B^2)) where I_0 is the
>0-th order modified Bessel function of the first kind.
>
>http://www.math.sfu.ca/~cbm/aands/page_376.htm

In fact it is 2pi I_0(sqrt(A^2 + B^2)), which can be computed as

oo
--- 2 pi A^2 + B^2 k
> ---- ( --------- )
--- k!^2 4
k=0

Rob Johnson <rob(a)trash.whim.org>
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