Friday, 9 October 2015

complex analysis - Show that $int _0 ^infty frac{x^{p-1}}{1+x} dx=frac{pi}{sin(p pi)}, 0




This is a complex analysis question, I tried to find the residue of the integrand at x=0, which equal to xp=0. So it looks the integral naively equal to 0. What did I do wrong?


Answer



basicly it comes from this theorm:



Theorom : 0xmxn+1dx=πnsin(m+1(πn)),nm2



Proof:



using contour integral we can write :

Czmzn+1dz



if C choosed ppropriately The integrand has n first-order singularities,at the n n-th roots of 1,these singular points are uniformly spaced around
the unit circle in the complex plane. thus using euler formula we can write:



1=ei(1+2k)π



singular points are located at:



zk=(1)1n=ei(1+2kn)π,k=0,1,2,3,...,n1




for other values of k these same n points simply repeat.



now focus on one of these singular points , k=0.



pick C to enclose just that one singularity, at z=z0=eiπn



enter image description here



so the central angle of the wedge is 2πn and the singularity is at half that πn.




contour’s three portions are:
C1=z=x,dz=dx,0xT



C2=z=Teiθ,dz=iTeiθdθ,0x2πn



C3=z=rei2πn,dz=ei2πndr,0rT



so:




Czmzn+1dz=T0xmxn+1dx+2πn0(Teiθ)m(Teiθ)n+1iTeiθdθ+0T(rei2πn)m(rei2πn)n+1ei2πndr



=T0xmxn+1dxT0rmei(m+1)2πnrn+1dr+2πn0Tm+1eimθTneinθ+1ieiθdθ



now clearly as T the θ-integral goes to zero because $m+1

T0rmei(m+1)2πnrn+1dr=ei(m+1)2πnT0xmxn+1dx



so as T:




Czmzn+1dz=0xmxn+1dx(1ei(m+1)2πn)



or as :



(1ei(m+1)2πn)=2isin((m1)πn)ei(m+1)2πn
we have:



Czmzn+1dz=2isin((m=1)πn)ei(m+1)2πn0xmxn+1dx



since we can write the integrand of the contour integral as a partial fraction

expansion:



zmzn+1=N0zz0+N1zz1+...+Nn1zzn1



where the N's are constants.integrating this expansion term-by-term:



Czmzn+1dz=N0Cdzzz0



using cauchy’s first integral theorem all the other integrals are zero ,so the only singularity C is z0 ,now using cauchy’s second integral theorem with f(z)=1 we get:




2isin((m=1)πn)ei(m+1)2πn0xmxn+1dx=2πiN0



and final step is calculate N0:



(zz0)zmzn+1=N0+N1(zz0)zz1+...



and if zz0 then:
N0=limzz0(zz0)zmzn+1=00
thus using L’Hoˆpital’s rule:
N0=limzz0(zz0)zmzn+1=zmn+10n




using z0=eiπn:
N0=ei(m+1n)πn
inserting this to result we finally get :
\bbox[5px,border:2px solid red] { \displaystyle \int_0^\infty {x^m\over{x^n+1}}dx={ {2\pi i}(-{e^{i( {m+1 \over n} )\pi} \over n }) \over {-2isin((m-1){\pi \over n})e^{i(m+1){2\pi \over n}}} } }
now in order to calculate what you asked:

define t=x^n then :
{dt \over dx} = nx^{n-1} so :
\int_0^\infty { t^{{m} \over n} \over t+1} ({dt \over nt^{{n-1}\over n} })=\int_0^\infty { t^{{{m+1} \over n}-1} \over {t+1}}dt



now define a={{m+1} \over n} and finally get($0 \bbox[5px,border:2px solid green] { \displaystyle \int_0^\infty {x^{a-1}\over{x+1}}dx= {\pi \over sin(a\pi)} }



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