Reply by Scott Hemphill April 3, 20102010-04-03
Tim Wescott <tim@seemywebsite.now> writes:

> OK: I'm posting this because I'm lazy. I haven't even _tried_ to > look this up yet, or to attempt to choke an answer out of Maxima. If > there _is_ an answer, I'm sure it'll be a Bessel function, and those > things always hit me like headlights do a deer. > > So, anyone feel like some math hand-holding? Here's the integral I'm > trying to find an answer for: > > integral from -pi to pi, e^(-a + b * cos(theta)) d theta, > > a > b > 0. > > This seems to be the class of problem that I can answer by myself, but > only after I've beat my head against the wall for ages or held myself > out for public ridicule -- I'm in a hurry, so I'm going to embarrass > myself _first_, in hopes of finding the solution quicker.
Mathematica 7.0: In[1]:= Integrate[E^(-a+b*Cos[theta]),{theta,-Pi,Pi},Assumptions->a>b>0] 2 Pi BesselI[0, b] Out[1]= ------------------ a E Scott -- Scott Hemphill hemphill@alumni.caltech.edu "This isn't flying. This is falling, with style." -- Buzz Lightyear
Reply by JCH April 1, 20102010-04-01
"glen herrmannsfeldt" <gah@ugcs.caltech.edu> schrieb im Newsbeitrag 
news:hp096e$blq$4@naig.caltech.edu...
> Tim Wescott <tim@seemywebsite.now> wrote: > >> So, anyone feel like some math hand-holding? Here's the integral I'm >> trying to find an answer for: > >> integral from -pi to pi, e^(-a + b * cos(theta)) d theta, > >> a > b > 0. > > Well, first I would factor out exp(-a), as that is constant. > > Next I put it into http://integrals.wolfram.com/ > > which says that "Mathematica could not find a formula for your > integral." > > But that only does indefinite integrals, and this seems like a > case where the definite integral might be possible to evaluate > even though the indefinte integral doesn't have an analytical > solution. (Not that you can find a numeric solution for an > indefinite integral.) >
>> integral from -pi to pi, e^(-a + b * cos(theta)) d theta,
For definite integral use Gaussian Integration: * http://home.arcor.de/janch/janch/_news/20100401-numerical-integration/ EXAMPLE: Page 1 PROGRAM: Page 2 -- Regards JCH u
> -- glen
Reply by Michael Plante March 31, 20102010-03-31
Glen wrote:
>Tim Wescott <tim@seemywebsite.now> wrote: > >> So, anyone feel like some math hand-holding? Here's the integral I'm >> trying to find an answer for: > >> integral from -pi to pi, e^(-a + b * cos(theta)) d theta, > >> a > b > 0. > >Well, first I would factor out exp(-a), as that is constant. > >Next I put it into http://integrals.wolfram.com/ > >which says that "Mathematica could not find a formula for your >integral." > >But that only does indefinite integrals, and this seems like a >case where the definite integral might be possible to evaluate >even though the indefinte integral doesn't have an analytical >solution. (Not that you can find a numeric solution for an >indefinite integral.)
Well, ignoring the exp(-a) as Glen did, I put it in Mathematica and obtained: Integrate[Exp[b*Cos[th]], {th, -pi, pi}] (where I've typed out "pi" for your newsreader's sake) And also ran: FullSimplify[%, Im[b] == 0 && Re[b] > 0] The result was: 2 pi (BesselI[0, b] - StruveL[0, b]) (...I'll post again if I see any font problems in google...) HTH, Michael
Reply by dvsarwate March 31, 20102010-03-31
On Mar 31, 2:24&#2013266080;pm, Tim Wescott <t...@seemywebsite.now> wrote:


> > integral from -pi to pi, e^(-a + b * cos(theta)) d theta,
Yes, it is a Bessel function (times a constant). Look at Abramowitz and Stegun, 9.6.16 which says that I_0(z) = (1/pi) integral from 0 to pi e^(z cos theta) d theta where I_0(z) is a modified Bessel function of the first kind and order 0. Shows up a lot in studies of noncoherent FSK, DPSK and the like and in the Ricean pdf which is the pdf of the amplitude of a sinusoid plus narrowband Gaussian noise. Hope this helps. --Dilip Sarwate
Reply by Jerry Avins March 31, 20102010-03-31
Tim Wescott wrote:
> OK: I'm posting this because I'm lazy. I haven't even _tried_ to look > this up yet, or to attempt to choke an answer out of Maxima. If there > _is_ an answer, I'm sure it'll be a Bessel function, and those things > always hit me like headlights do a deer. > > So, anyone feel like some math hand-holding? Here's the integral I'm > trying to find an answer for: > > integral from -pi to pi, e^(-a + b * cos(theta)) d theta, > > a > b > 0. > > This seems to be the class of problem that I can answer by myself, but > only after I've beat my head against the wall for ages or held myself > out for public ridicule -- I'm in a hurry, so I'm going to embarrass > myself _first_, in hopes of finding the solution quicker.
I'm too rusty to see why iterated integration by parts won't work. Jerry -- Discovery consists of seeing what everybody has seen, and thinking what nobody has thought. .. Albert Szent-Gyorgi &#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
Reply by glen herrmannsfeldt March 31, 20102010-03-31
Tim Wescott <tim@seemywebsite.now> wrote:
 
> So, anyone feel like some math hand-holding? Here's the integral I'm > trying to find an answer for:
> integral from -pi to pi, e^(-a + b * cos(theta)) d theta,
> a > b > 0.
Well, first I would factor out exp(-a), as that is constant. Next I put it into http://integrals.wolfram.com/ which says that "Mathematica could not find a formula for your integral." But that only does indefinite integrals, and this seems like a case where the definite integral might be possible to evaluate even though the indefinte integral doesn't have an analytical solution. (Not that you can find a numeric solution for an indefinite integral.) -- glen
Reply by Tim Wescott March 31, 20102010-03-31
Tim Wescott wrote:
> OK: I'm posting this because I'm lazy. I haven't even _tried_ to look > this up yet, or to attempt to choke an answer out of Maxima. If there > _is_ an answer, I'm sure it'll be a Bessel function, and those things > always hit me like headlights do a deer. > > So, anyone feel like some math hand-holding? Here's the integral I'm > trying to find an answer for: > > integral from -pi to pi, e^(-a + b * cos(theta)) d theta, > > a > b > 0. > > This seems to be the class of problem that I can answer by myself, but > only after I've beat my head against the wall for ages or held myself > out for public ridicule -- I'm in a hurry, so I'm going to embarrass > myself _first_, in hopes of finding the solution quicker. >
Worse, maybe it's _not_ covered by Bessel functions, and I'm _really_ on my own. Argh. (Proceeding to plug away at the math). -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
Reply by Tim Wescott March 31, 20102010-03-31
OK:  I'm posting this because I'm lazy.  I haven't even _tried_ to look 
this up yet, or to attempt to choke an answer out of Maxima.  If there 
_is_ an answer, I'm sure it'll be a Bessel function, and those things 
always hit me like headlights do a deer.

So, anyone feel like some math hand-holding?  Here's the integral I'm 
trying to find an answer for:

integral from -pi to pi, e^(-a + b * cos(theta)) d theta,

a > b > 0.

This seems to be the class of problem that I can answer by myself, but 
only after I've beat my head against the wall for ages or held myself 
out for public ridicule -- I'm in a hurry, so I'm going to embarrass 
myself _first_, in hopes of finding the solution quicker.

-- 
Tim Wescott
Control system and signal processing consulting
www.wescottdesign.com