Nice stack exchange steal.
Haha was redirected there by someone after I couldn't answer it for themNice stack exchange steal.
Observe that the term
By inspection, the answer to the first integral is:
The answer to this integral is:
Can you please show working out? I can't really learn from this and what is this "slight manipulation" that you have stated.By inspection, the answer to the first integral is:
Similarly, the answer to the second integral is:
These come from the integral definitions of the Riemann Zeta and Dirichlet Eta functions, respectively, with slight manipulation.
The function to consider is:Evaluate:
Can you please show working out? I can't really learn from this and what is this "slight manipulation" that you have stated.
anyway let's bump this one, been sitting here for a very long time.Evaluate:
Hence, or otherwise, evaluate:
Beta Function manipulation is pretty trivial....
Beta function, second derivative with respect to its parameter to get rid of the log squared term, beta function reflection formula is then related to the digamma function (and of course its second derivatives will be needed along the way), then a reflection formula for the digamma function and I guess we are home and hosed but is it not nice to see something differentBeta Function manipulation is pretty trivial....