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ò₀ ^¥[1/(1+xⁿ)]dx = p/(nsin(p/n))

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By Arun Iyer on Wednesday, December 19, 2001 - 06:10 pm:

how to prove this thing.....

ò₀ ^¥[1/(1+xⁿ)]dx = [p/(nsin(p/n))]

love arun

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By Michael Doré on Thursday, December 20, 2001 - 02:01 am:

For n > 1 I think this is doable by residues. Try integrating round a semi-circle of radius T in the Argand diagram, centre origin with diameter along the real line, and with the arc in the half of the plane with positive imaginary part. Apply Cauchy's second theorem and finally take the limit as T->¥.

For n £ 1 the integral clearly diverges.

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By Arun Iyer on Thursday, December 20, 2001 - 06:05 pm:

sorry,i forgot to mention the condition that n>1

anyways to be quite frank with you...i am not quite well versed with complex integration..

i know what complex integration is,the cauchy riemann conditions,cauchy residue theorem.....and a bit about branch cuts..

however,i have not yet reached a stage where i can solve a problem of complex integration...

so can you give me any other way of proving this(if possible) OR give me the complete solution of complex integration.....(i will try my best to understand..)

love arun

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By Kerwin Hui on Thursday, December 20, 2001 - 11:31 pm:

Alternatively try to contour integrate (1+zⁿ)^-1 along the sector G consists of
C₁={x: x from 0 to R}
C₂={R e^iq: q from 0 to 2p/n}
C₃={ye^2ip/n: y from R to 0}

This has the advantage that we only have one pole to consider.

The only pole in the region bounded by G is at e^ip/n, which has residue -n^-1e^ip/n.

It is easy to check that the integral along C₂ tends to 0 as R®¥, and the integral along C₃ is just -e^2ip/n times the integral along C₁, our required integral. Hence

ò₀ ^¥(1+xⁿ)^-1dx=[1-e ^2pi/n]^-1 2pi (-n^-1e^ip/n)=p/[n sin(p/n)].

Kerwin

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By Arun Iyer on Friday, December 21, 2001 - 06:33 pm:

i repeat again....i am quite weak(Actually very weak)in contour integration....

hence a detailed solution would be quite helpful...i am quite sorry for all the trouble i am causing....

love arun

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By Kerwin Hui on Saturday, December 22, 2001 - 06:17 am:

Arun,

There are various ways in which we can do a real integral by complex integration. In your case, since one limit is infinity, we do the following:

1. Change x to z
2. Choose our contour G such that we have some part resembling the integral we want.
3. Do the integral in two ways and equate the two results to find the answer.

So here is the complete solution:

We will consider the integral ò_G (1+zⁿ)^-1dz, where G is the contour stated in my previous post.

First way to do this integral is by Cauchy's residue theorem. The only pole inside our region is at e^ip/n (providing we choose R>1, which is true since we are going to take the limit R®¥). It is straightforward to check the residue at this pole is -e^ip/n/n. Hence by Cauchy's residue theorem, we have

ò_G (1+zⁿ)^-1dz = -2pie^ip/n/n

Now we do the integral by another method. Splitting the contour into C₁,C₂ and C₃:

ò_{C_1} (1+zⁿ)^-1dz = ò₀ ^R(1+xⁿ)^-1dx
ò_{C_2} (1+zⁿ)^-1dz = ò₀ ^2p/n [1+(Re^iq)ⁿ]^-1 iR e^iq dq
ò_{C_3} (1+zⁿ)^-1dz = ò_R ⁰[1+(ye^2pi/n)ⁿ]^-1 e^2pi/n dy = -e^2pi/n ò₀ ^R(1+yⁿ)^-1 dy

Hence, for all R>1,
(1-e^2pi/n)ò₀ ^R(1+xⁿ)^-1dx + iò₀ ^2p/nR[1+Rⁿe^{in q}]^-1e^iqdq = -2pie^ip/n/n

Now we let R®¥. Can you see why the q integral vanish? The final step is to invoke the definition of sine (sin z=(e^iz-e^-iz)/(2i)) and the answer comes out quite nicely.

Kerwin

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