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From: Robert Israel on 23 Jun 2010 22:41
Dan Cass <dcass(a)sjfc.edu> writes:

> > 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)).

True: that is not an elementary function.

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

This one the latest version (Maple 14) does know, although previous versions
didn't. On the other hand int(exp(sin(x)), x = 0 .. 2*Pi) still returns
unevaluated.

Of course a change of variables makes Han's integral into
int(exp(R*cos(t)),t=0..2*Pi) = 2*Pi*I_0(R)
(where R = sqrt(A^2 + B^2), and I_0 is the modified Bessel function of the
first kind with order 0).


> This may mean that even a complex analysis contour integral approach will
> not work.

That is not particularly relevant one way or the other.

In this case one way to see the result is to expand the exponential as a
power series, note that
int(cos(t)^n, t=0..2*Pi) = 0 if n is odd (by symmetry)
= 2 pi n!/(((n/2)!)^2 2^n) if n is even
(by writing cos(t) = (exp(it) + exp(-it))/2 and expanding),
and use I_0(x) = sum(4^(-k)/k!^2*x^(2*k),k = 0 .. infinity).
Another way is to differentiate the integral twice and use integration
by parts to show that the integral satisfies the modified Bessel differential
equation of order 0.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Han de Bruijn on 24 Jun 2010 05:03
On Jun 23, 5:47 pm, r...(a)trash.whim.org (Rob Johnson) wrote:
> In article <c641a0f7-31f3-4fe7-b393-b19339c2c...(a)5g2000yqz.googlegroups.com>,
>
> dvsarwate <dvsarw...(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 <r...(a)trash.whim.org>
> take out the trash before replying
> to view any ASCII art, display article in a monospaced font

And only six (6) terms of that series are more than sufficient. Thanx!

Han de Bruijn
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