areslagae

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These are answers submitted by areslagae

The integral

assume(a > 0);
int(exp(a*cos(phi))*sin(phi)^2, phi = 0 .. Pi);

equals to

(Pi/a)*BesselI(1,a)

I have solved the integral using Mathematica, which seems to solve all these integrals out of the box.

 

I am having troubles with another similar integral.

 

assume(a > 0);
int(exp(a*cos(phi))*sin(phi)^2, phi = 0 .. Pi);
 

Jean-Marc,

 

Thanks a lot!

 

Do you have any idea how to compute the integral by hand? I have checked Abramowitz and Stegun, but could not find anything of use.

 

I have tried the free web service, but it seems to be very limited.

Jean-Marc,

Thanks for your quick reply.

 

I am actually trying to solve

int(int(exp(a*cos(theta)*sin(phi))*sin(phi), phi = 0 .. Pi), theta = 0 .. 2*Pi) (1)

with a a constant.

Using the second last definite integral on

http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions

the integral (1) reduces to

2*Pi*(int(sin(phi)*BesselI(0, a*sin(phi)), phi = 0 .. Pi)) (2)

if I am not mistaken.

(By the way, why does mathematica not seem to know this definite integral?)

 

Does the above mean that mathematica can solve the integral (2)? If so, would you please be so kind to try? I am not familiar with "HoldForm" and the "=="

 

Is there an online version of Mathematica that can solve these integrals?

 

 

Thanks for your reply. This definitely is a different Hankel transform, but still not the standard one. With the example above: > hankel:=(f,r,k,nu)->inttrans[hankel](f*r^(1/2),r,k,nu)/k^(1/2): > hankel((1/2)*Dirac(r-a),r,k,0); 3/2 a BesselJ(0, k a) 1/2 -------------------- 1/2 k instead of the desired Pi BesselJ(0, 2 Pi a q) a Best regards,

Thank you for your reply.

I also tried what you suggested, and it works only for some transforms.

In some cases, the inverse transform cannot be computed when the forward transform can be computed.

For example, using a pair in table 13.2 of Bracewell (p 338):

> restart;
> assume(a > 0);
> f(r):=1/(2)Dirac(r-a);
     1            
r -> - Dirac(r - a)
     2            
> 2*Pi*(int(f(r)*BesselJ(0, 2*Pi*r*q)*r, r = 0 .. infinity));
                          Pi BesselJ(0, 2 Pi a q) a
> f(r):=Pi*BesselJ(0,2*Pi*a*r)*a;
r -> Pi BesselJ(0, 2 Pi a r) a
> 2*Pi*(int(f(r)*BesselJ(0, 2*Pi*r*q)*r, r = 0 .. infinity));
                                Pi undefined
 

I was under the impression that inttrans/hankel uses a table of known transforms and would be able to compute transforms that cannot be computed with int(...). That motivates my original question.

Is there any way to compute the hankel transform I want using the extra knowledge of inttrans/hankel?

Why does Maple use an other definition?

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