Sunday, 10 July 2016

calculus - Is there a general formula for I(m,n)?




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)=2π33 I(1,4)=π4 I(2,4)=π22



For natural m,n the integral exists if and only if nm+2.




Is there a general formula for I(m,n) with integers m,n and 0mn2 ?




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+1iReiϕdϕR0xmei2πm/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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