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Contour integral
By Kerwin Hui on Tuesday, June 8, 1999 - 07:07 pm:

How do you prove that


ó
õ		 ¥

0 
  (lnx)2
1+x2
 dx =  p3
8
 ?
By Alex Barnard (Agb21) on Monday, June 14, 1999 - 02:57 pm:

The best way to show this is to use complex functions. But before writing out an answer can I check how much (if at all) you know about them.

Do you know what a complex number is?
Do you know what a residue of a function is?
Have you heard of the calculus of residues?

Let me know how much of these you know about and then I'll know how detailed to make my answer.

Many thanks,

AlexB.

By Kerwin Hui (P1312) on Thursday, October 21, 1999 - 06:31 pm:

I do know something about complex number and only a little about residues. Thanks.

By Alex Barnard (Agb21) on Tuesday, October 26, 1999 - 11:07 am:

Firstly you need to rearrange the integral a little bit. Make the substitution x=ey. You should find that the integrand becomes much nicer. I'll leave you to work out what it simplifies to (but you should find it involves a COSH term on the denominator).

Do you know about hyperbolic functions?

Do you know where there zeros are (you need this to work out where the residues are for the integral you want)?

Do you know that they are periodic? What is their period?

Let me know what you get for all of these and I'll carry on.

AlexB.

By Kerwin Hui (P1312) on Tuesday, October 26, 1999 - 12:23 pm:

Thanks, Alex.

Well, I've used the substitution given to get the integrand as
y2/(2cosh y)
with the limits from -infinity to +infinity.

The denominator is a cosh function, so the period is 2ip

The zeros will be where cosh y=0, so Im(y)=p(2n+1)/2

But the real part of y can be any real number??? Help!!!

By Alex Barnard (Agb21) on Tuesday, October 26, 1999 - 01:06 pm:

Well that is all correct. Now all you have to do is to pick the correct contour to integrate around. There is one which is very common to use for periodic functions. It is a rectangle with the short side being the half-period.

In our case imagine the rectangle with vertices at the following points:

r+0i, r+pi, -r+pi, -r+0i.

Where I am taking the route round the contour exactly as I have written the points.

And we are going to integrate the function z2/cosh z (ignore the 2 for now - just remember to put it back later!!) around the rectangle contour.

Can you work out the value of this integral by contour integration (you have already worked out where the poles are - just work out the residues too)?

Can you see what the value of the function is along the two horizontal sides (they should look very similar!)?

Can you see that the value of the integral along the two vertical sides decreases very quickly as r increases?

So can you work out how the contour integral is related to the integral you wanted?

Let me know how you get on with these...

AlexB.

By Kerwin Hui (P1312) on Tuesday, October 26, 1999 - 02:06 pm:

Thanks, I got stuck at exactly the same point as before, namely, the infinite number of zeros. How can I get round the problem?

By Alex Barnard (Agb21) on Tuesday, October 26, 1999 - 04:54 pm:

There is no problem. There are only a finite number of zeros inside the contour I chose.

By Kerwin Hui (P1312) on Saturday, October 30, 1999 - 09:51 pm:

I see what you mean now. I made the mistake above thinking that the real of y can be anything, but obviously it has to be zero to reduce to cos(Im(y)).

I found the contour integral to be

-p3/2

But I can't see what the values along the sides of rectangle.

By Alex Barnard (Agb21) on Monday, November 1, 1999 - 12:50 pm:

Okay... that is the correct value for the contour integral. Now we must look at the functions we get along each of the sides. From this we will be able to work out the integral you wanted.

Along the bottom side we have x2/cosh(x) [which is what we are looking for].

Along the top side we have (x+ip)2/cosh(x+ip) and this simplifies to:
(x2 + 2ipx - pi2) /-cosh(x)

Along the left side we have (r+iy)2/cosh(r+iy)

Along the right side we have (-r+iy)2/cosh(-r+iy)

And we know that adding up all the integrals of these will give us -p3/2. You can see some copies of the integral you want on the two horizontal sides (with some extra terms). Now that you know the function along the left and right sides can you show that they get very small as r gets large. And hence that as r tends to infinity they give zero contribution to the overall integral?

AlexB.

By Kerwin Hui (P1312) on Monday, November 1, 1999 - 06:01 pm:

Thanks, Alex.

I can see that the two vertical sides gives zero contribution to the contour integral. However, the problem is with the two horizontal sides, giving the integral to be a negative value. I know it must be trivial sign error but I just cannot see how to fix it.

By Alex Barnard (Agb21) on Wednesday, November 3, 1999 - 03:07 pm:

Ignoring the two vertical sides we get just the terms from the horizontal sides. The bottom one gives us the integral we want (call it A). The top side gives us (remember that this integral in done in the backwards direction):

A - Integral[p2/cosh(y)] + something imaginary

So in total we get:

2A - B + Ci = -p3 / 2

where B is the integral and C is a real number.

So now we have to work out B. This can be done by an identical method to everything above (or by knowing the answer!!).

B = p3.

So 2A + Ci = p3 / 2

Now C must be 0 because A is real and p3 is real.

Hence A = p3 / 4.

And if you remember I dropped a factor of 2 a few messages ago and so you get p3 / 8 in the end!

AlexB.

By Kerwin Hui (P1312) on Wednesday, November 3, 1999 - 03:23 pm:

Thanks, Alex.

I post this message after I complete the solution myself. I see what was wrong - I've forgotten the sech x term.

òsech x dx = arctan(sinh x) + constant, or equivalently
2*arctan(ex) + constant, of course.