I've been working on this for the past hour.

let

then

let

then

which implies that

so finally ?

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- August 19th 2009, 09:24 PMRandom VariableIs this integral correct?
I've been working on this for the past hour.

let

then

let

then

which implies that

so finally ? - August 19th 2009, 10:38 PMCalculus26
Its correct according to Mathcad

- August 19th 2009, 10:51 PMRandom Variable
- August 19th 2009, 10:53 PMCalculus26
Mathcad does the symbolic calculation for arbitrary a-- its ruining my skills

- August 20th 2009, 12:41 AMBruno J.
Do you know the residue theorem? I haven't checked fully but it seems that such an integral would yield fairly easily to it.

- August 20th 2009, 04:42 AMMoo
Hello,

As Bruno J. mentioned, it can be done with the residue theorem.

If you consider the keyhole contour, then it can be proved that :

where the branch of the logarithm is such that

It's not more beautiful than your method though. - August 20th 2009, 10:05 AMRandom Variable
I didn't even think about using contour integration. The problem is actually an exercise in my complex analysis textbook.

Moo suggested using the keyhole contour. My book suggests using a semicircle in the upper-half plane with a little semicircle about the origin (because ln z is undefined for z=0). They also suggest using the branch so that ln z is analytic on and inside of the contour.

So let R be the radius of the big semicircle and r be the radius of the small semicircle

and let

then

so there is a simple pole inside of the contour at

then

So I have

now letting R go to infinity and r go to zero

I'm just assuming that the two contour integrals evaluate to zero. I'll try to offer a justification in another post. But I'll probably need some help.

Equating real parts on both sides I have

so finally

(Whew) - August 20th 2009, 11:16 AMRandom Variable
nothing to see here

- August 20th 2009, 06:08 PMRandom Variable
OK. After a bit of confusion about the reverse triangle inequality, I think I can show that

let

then

which according to the ML inequality shows that

The other one is similar except r<a.