Consider the integral
I(m,n):=∫∞0xmxn+1dx
For m=0, a general formula is I(0,n)=πnsin(πn)
Some other values are I(1,3)=2π3√3 I(1,4)=π4 I(2,4)=π2√2
For natural m,n the integral exists if and only if n≥m+2.
Is there a general formula for I(m,n) with integers m,n and 0≤m≤n−2 ?
Answer
We can use contour integration to arrive at the general result. Note that
∮Czmzn+1dz=2πiRes(zmzn+1,z=eiπ/n)=−2πieiπ(m+1)/nn
where C is the "pie slice" contour comprised of (i) the real-line segment from 0 to R, where R>1, (ii) the circular arc of radius R that begins at R and ends at Rei2π/n, and (3) the straight line segment from Rei2π/n to 0.
Then, we can write
∮Czmzn+1dz=∫R0xmxn+1dx+∫2π/20RmeimϕRneinϕ+1iReiϕdϕ−∫R0xmei2πm/nxn+1ei2π/ndx
If n>m+1, then as R→∞, the second integral on the right-hand side of (2) vanishes and we find that
\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{x^m}{x^n+1}\,dx=2\pi i\frac{e^{i\pi(m+1)/n}}{n(e^{i2\pi(m+1)/n}-1)}=\frac{\pi/n}{\sin(\pi(m+1)/n)}}
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