ò0¥ x2/(x4+1) dx=p/(2Ö2)
By Georgina Horton on Monday, August 26,
2002 - 01:32 pm:
Hello. Can someone please show me how to prove the following using a
suitable contour integral:
òx2/(x4+1) dx=p/(2Ö2) with limits 0 and ¥
By Andre Rzym on Tuesday, August 27,
2002 - 07:52 pm:
First of all, let's see where the poles/essential singularities are. If you
think about the denominator, it will be zero when x2 = ±i, so using
partial fractions
x2 /(x4 +1)=1/2[1/(x2 +i)+1/(x2 -i)]
The two denominators on the rhs will be zero when
; x=±Öi respectively, so we can expand again:
x2 /(x4 +1)=1/4[(1/a)1/(x-a)-1/(x+a)+(1/b)1/(x-b) -1/(x+b)]
where
a = -1/Ö2+i/Ö2 b = 1/Ö2+i/Ö2
Now consider your integral but on a closed contour being a semicircle, centred
on the origin, with the 'flat side' being on the real axis, and the curved side
being above the real axis.
You need to convince yourself:
(i) That as the radius tends to infinity, the integral vanishes over
the 'curved side'.
(ii) The integral over the 'flat side' is twice your integral as the
radius -> infinity
(iii) The first and third fractions in the equation above produce
simple poles enclosed within the semicircle. The residues are 1/4a,
1/4b respectively
So
| òC x2 /(x4 +1)dx=2 ò0¥x2 /(x4 +1)dx=2 pi |
å
|
|
where C is the semicircle contour (in the limit as its radius -> infinity)
and
is the sum of the residues, i.e.
So
ò0¥x2 /(x4 +1)dx = pi.-i/2Ö2=p/2Ö2
Andre