Here's a "proof" I came up with when asked to use this integral for a physics homework problem in undergrad (I think we were expected to cite wolfram alpha). Perhaps worth adding for the novelty, if not the rigor, I think: Define I(a)=∫∞0sin(2πax)xdx=i2∫...
Edit: By letting sinx2√=tsinx2=t We get: I=2∫12√0arcsinx12−x2−−−−−−√dx=2Li2(12–√)−π224+ln224I=2∫012arcsinx12−x2dx=2Li2(12)−π224+ln224 Where the latter integral was evaluated with wolfram. I would love to see a proof ...
The speaker has tried using Mathematica and WolframAlpha but the solutions given are too complex. They suggest using the Weierstraß substitution to simplify the integral, but even then, the solution is still complicated. The speaker also mentions trying an integration by parts method, but it ...
V. The Direct Numerical Calculation of Elliptic Functions and Integrals. London: Cambridge University Press, 1924.Prasolov, V. and Solovyev, Y. Elliptic Functions and Elliptic Integrals. Providence, RI: Amer. Math. Soc., 1997.Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and ...