I am trying to calculate the following principle value integral
P∫∞0xλ−11−xdx
for λ∈[0,1]. I tried to turn this into a contour integral so our complex function is given by
f(z)=zλ−11−z
which has a simple pole at z=1 and branch points at z=0 and z=∞. We integrate over a contour with two indented paths thus we pick up half residues at these two contours), thus we can write the contour C as
C=6∑i=1Ci+Cϵ+CR=C1+C2+C3+C4+C5+C6+Cϵ+CR.
Since the contour encloses no poles and f(z) is holomorphic, by the Cauchy-Goursat theorem we know that
∮Cf(z)dz=0.
I can show that the integrals of the contours CR and Cϵ vanish since
|∫CR|≤|∫2π0dθRλ−1RR|=|2πR1−λ|→0 as R→∞ for λ<1
and
|∫Cϵ|≤|∫2π0dθϵλ−1⋅ϵ=|2πϵλ|→0 as ϵ→0 for λ>0.
Now write the contour integral as
0=∮Cf(z)dz=P∫C1+ P∫C2+ P∫C3+ P∫C4+ P∫C5+ P∫C6.
Explicitly we can now calculate three contour integrals over C1,C2,C3 by using z=xe2πi, dz=dxei2π=dx and we obtain
P∫C1+ P∫C2+ P∫C3=lim
Note the first integral in terms of z can just be written as
\lim_{R\to\infty} \lim_{\epsilon \to 0}\int_{Re^{i(2\pi-\epsilon)}}^{(1+\epsilon)2\pi i} \frac{z^{\lambda-1}}{1-z}dz=\int_{\infty}^{1} \frac{x^{\lambda-1}}{1-x}dx
however I am stuck as to how go from here. Thanks
Answer
I'll evaluate the more general case \text{PV} \int_{0}^{\infty} \frac{x^{\lambda-1}}{x^{b}-1} \ dx \ \ (b >\lambda > 0, \ b\ge 1) .
Let \displaystyle f(z) = \frac{z^{\lambda-1}}{z^{b}-1}, where the branch cut is along the positive real axis.
Now integrate around a wedge of radius R that makes an angle of \displaystyle \frac{2 \pi }{b} with the positive real axis and is indented around the simple poles at z=1 and z=e^{2 \pi i /b}, and the branch point at z=0.
The integral obviously vanishes along the arc of the wedge as R \to \infty.
And there is no contribution from the indentation around the branch point at z=0 since \Big| \int_{0}^{\frac{2 \pi}{b}} f(re^{it}) ire^{it} \ dt \Big| \le \frac{2 \pi}{b} \frac{r^{\lambda}}{1-r^{b}} \to 0 \ \text{as} \ r \to 0.
Then going around the contour counterclockwise,
\text{PV} \int_{0}^{\infty} f(x) \ dx - \pi i \ \text{Res} [f, \pi i] + \ \text{PV}\int_{\infty}^{0} f(te^{\frac{2 \pi i }{b}}) e^{\frac{2 \pi i}{b}} \ dt - \pi i \ \text{Res}[f,e^{\frac{2 \pi i}{b}}] = 0 .
Looking at each part separately,
\text{Res}[f,1] = \lim_{z \to 1} \frac{z^{\lambda-1}}{bz^{b-1}} = \frac{1}{b}
\text{PV} \int_{\infty}^{0} f(te^{\frac{2 \pi i }{b}}) e^{\frac{2 \pi i}{b}} \ dt = - e^{\frac{2 \pi i}{b}} \text{PV} \int_{0}^{\infty} \frac{t^{\lambda-1} e^{\frac{2 \pi i(\lambda-1)}{b}}}{t^{b} e^{2 \pi i} - 1} \ dt = - e^{\frac{2 \pi i \lambda}{b}} \text{PV} \int_{0}^{\infty} \frac{t^{\lambda-1}}{t^{b}-1} \ dt
\text{Res}[f, e^{\frac{2 \pi i}{b}}] = \lim_{z \to e^{\frac{2 \pi i}{b}}} \frac{z^{\lambda-1}}{bz^{b-1}} = \frac{e^{\frac{2 \pi i (\lambda-1)}{b}}}{b e^{\frac{2 \pi i(b-1)}{b}}} = \frac{1}{b} e^{\frac{2 \pi i \lambda }{b}}
Plugging back in and rearranging,
\text{PV} \int_{0}^{\infty} \frac{x^{\lambda-1}}{x^{b}-1} \ dx = \frac{\pi i}{b} \frac{1 + e^{\frac{2 \pi i \lambda}{b}}}{1-e^{\frac{2 \pi i \lambda}{b}}} = - \frac{\pi}{b} \cot \left(\frac{\pi \lambda}{b} \right)
or
\text{PV} \int_{0}^{\infty} \frac{x^{\lambda-1}}{1-x^{b}} \ dx = \frac{\pi}{b} \cot \left(\frac{\pi \lambda}{b} \right)
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