Arun Iyer
Posted on Monday, 09 June, 2003 - 10:51 am:
Using contour integration, evaluate


ó
õ 		¥

0 
 [(sinhax)/(sinhpx)]dx
-p < a < p

I am not able to think of a proper contour for this.

love arun

David Loeffler
Posted on Monday, 09 June, 2003 - 10:54 am:
Try using a long thin rectangle with vertices ±R, ±R + i.

David
Arun Iyer
Posted on Monday, 09 June, 2003 - 12:55 pm:
Thanks David!
i got close to the answer but not very close.
i got the answer as 2tan(a/2) but my text says (1/a)*tan(a/2)

I will put my working here.

This is the contour ...
contour
(Sorry for the shabby drawing)
Now since the pole z=0 is on real axis,i had to indent the contour by drawing a semi-circle.

 The only singularity then is i which is on the contour,

R(i)=(-i/p)sin(a)

so the integral along the contour =2sin(a)

Thus the given integral breaks up into six integrals,

ò-R-r+ò-rr+òrR+òAB+ò BC+òCD

if we let r® 0 and R®¥, the second, fourth and sixth integrals are zero (if I have not made any mistake).

The first and third then get merged.

Further simplification gives,

(1+cos(a))ò-¥¥[sinh(ax)/sinh(px) ]dx+isin(a)ò-¥¥ [cosh(ax)/sinh(p x)]dx

Then the comparison of real and imaginary parts gives me the answer.

What mistake have I made?

love arun

Kerwin Hui
Posted on Monday, 09 June, 2003 - 07:04 pm:
I get (1/2)tan (a/2) as the answer. You can actually have your contour passing through 0, since the singularity at 0 is removable.

There are a couple of mistakes in your calculation:
1. You cannot have your contour passing through a pole. You need to take a small semicircular arc about the point i - and this path has nontrivial contribution to the integral.
2. The question asks for the integral over the nonnegative reals, not the whole real line.

Kerwin
Arun Iyer
Posted on Tuesday, 10 June, 2003 - 07:20 am:
When I had to evaluate ò0¥ sinz/z I had to indent the contour at z=0 to evaluate the answer even though it was removable???!!!???

and yes!! i forgot it was over the non-negative reals not the whole real line

love arun
Demetres Christofides
Posted on Wednesday, 11 June, 2003 - 01:53 pm:
Arun, when you want to evaluate ò0¥ sinx/x dx, you integrate eiz/z over an appropriate contour. This has a pole at 0.

Demetres

Arun Iyer
Posted on Thursday, 12 June, 2003 - 02:33 pm:
Oops Sorry!!
Thanks Demetres.(Thanks Kerwin and David)

Another Question,
How to do ....
 ò0¥(sinx/x)2 dx

love arun

Kerwin Hui
Posted on Thursday, 12 June, 2003 - 04:48 pm:
Integrate by parts gives
LaTeX Image

Kerwin
Arun Iyer
Posted on Thursday, 12 June, 2003 - 05:36 pm:
Smoooooooth!!
Thanks Again!

love arun