Find the residue in of .
My attempt : I notice that is analytic except in . (well I think so! I guess I should show that is analytic. Is that sufficient that f satisfy C-R equations? Because as you'll see, I need the analyticity of for the coming part)
I believe that .
I forgot something very important : why does ? I mean, I think there's a theorem that justify it (it seems it's not Cauchy-Goursat theorem) but I forgot which one.
Am I right?
The powers of z in the double sum have the form . We want all the 1/z terms or when or when and when n is that value, the coefficient on 1/z is . I'm just replacing the (n-k) expression in the double sum with (2k+1-k). But since k is going from 0 to n and n is going to infinity, that's the same as k going from 0 to infinity (eventually) and that's why I use the summation expression.
Try and work through the same process for the following:
Very good Arbolis. I think you have it now. Here's the numeric check in Mathematica which expresses the sum in terms of some obscure KevinBel function.
Code:In[208]:= Sum[(-1)^k/((2*k)!*(2*k + 1)!), {k, 0, Infinity}] NIntegrate[Cos[z]*Exp[1/z]*I*Exp[I*t] /. z -> Exp[I*t], {t, 0, 2*Pi}] N[2*Pi*I*((KelvinBei[1, 2] - KelvinBer[1, 2])/Sqrt[2])] Out[209]= (KelvinBei[1, 2] - KelvinBer[1, 2])/ Sqrt[2] Out[210]= -4.440892098500626*^-16 + 5.761766462098244*I Out[211]= 0. + 5.7617664620982385*I
Of related interest:
Wolfram|Alpha 1
Wolfram|Alpha 2