∫∞0sin(2πx)x(x2+3)
I looked at e2πizz3+3z, also calculated the residues, but they don't get me the right answer. I used that ∫∞−∞f(z)dz=2πi(∑Reszr)+πiRes0, but my answer turns out wrong when I check with wolframalpha.
Residue for 0 is 1, for z=√3i it's −e−2π2 . . .
In a worse attempt I forgot 2π and used z only (i.e. eizz3+3z) and the result was a little closer, but missing a factor of 2 and and i.
Can anyone see the right way? Please do tell.
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