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)),n−m≥2
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,...,n−1
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
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,0≤x≤T
C2=z=Teiθ,dz=iTeiθdθ,0≤x≤2πn
C3=z=rei2πn,dz=ei2πndr,0≤r≤T
so:
∮Czmzn+1dz=∫T0xmxn+1dx+∫2πn0(Teiθ)m(Teiθ)n+1iTeiθdθ+∫0T(rei2πn)m(rei2πn)n+1ei2πndr
=∫T0xmxn+1dx−∫T0rmei(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)2πn∫T0xmxn+1dx
so as T→∞:
∮Czmzn+1dz=∫∞0xmxn+1dx(1−ei(m+1)2πn)
or as :
(1−ei(m+1)2πn)=−2isin((m−1)πn)ei(m+1)2πn
we have:
∮Czmzn+1dz=−2isin((m=1)πn)ei(m+1)2πn∫∞0xmxn+1dx
since we can write the integrand of the contour integral as a partial fraction
expansion:
zmzn+1=N0z−z0+N1z−z1+...+Nn−1z−zn−1
where the N's are constants.integrating this expansion term-by-term:
∮Czmzn+1dz=N0∮Cdzz−z0
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)2πn∫∞0xmxn+1dx=2πiN0
and final step is calculate N0:
(z−z0)zmzn+1=N0+N1(z−z0)z−z1+...
and if z→z0 then:
N0=limz→z0(z−z0)zmzn+1=00
thus using L’Hoˆpital’s rule:
N0=limz→z0(z−z0)zmzn+1=zm−n+10n
using z0=eiπn:
N0=−ei(m+1n)πn
inserting this to result we finally get :
∫∞0xmxn+1dx=2πi(−ei(m+1n)πn)−2isin((m−1)πn)ei(m+1)2πn
now in order to calculate what you asked:
define t=xn then :
dtdx=nxn−1 so :
∫∞0tmnt+1(dtntn−1n)=∫∞0tm+1n−1t+1dt
now define a=m+1n and finally get($0∫∞0xa−1x+1dx=πsin(aπ)
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