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 Oct 21 awarded Yearling May 30 asked Non trivial integral with the Bose-Einstein distribution and Cosine function Feb 25 asked Differentiation under the double integral sign Sep 19 answered Transforming a sum in an integral Sep 13 awarded Editor Sep 13 revised Transforming a sum in an integral added 4 characters in body Sep 13 revised Transforming a sum in an integral edited tags Sep 13 answered Transforming a sum in an integral Sep 12 comment Indefinite integral $\int^{\infty}_{0}\frac{x}{x^4+1}dx$ via residues Sep 12 awarded Teacher Sep 12 answered Why does this expression equal $\pi$? Sep 12 comment How to justify term-by-term expansion to compute an integral Sorry but i don't know how to calculate this integral by series. I hope this will be useful for you ,despite the fact it doesn't use series Sep 12 answered How to justify term-by-term expansion to compute an integral Sep 11 asked Transforming a sum in an integral Sep 11 comment Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ I did the calculations with derive!It's all right! Sep 11 comment Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ I don't know how to handle this integral. I think he means that $v' = \frac{{\sqrt 2 }}{2}v - \frac{{\sqrt 2 }}{2}w$ and $w' = \frac{{\sqrt 2 }}{2}v + \frac{{\sqrt 2 }}{2}w$. The derivate is of v' now. If you do the calculations you get the correct answer. Sep 11 comment Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ Can you explain me the formula of quadratic form?If there are other threads about this thing please send me to these. Sep 11 awarded Supporter Sep 11 awarded Nice Question Sep 11 comment Evaluating the integral $I(u,v,w)=\iint_{(0,\infty)^2}\sinh(upq) e^{-vq^2 - wp^2}pq(q^2-p^2)^{-1}dpdq$ i want to tell you that there is a mistake. THe quantity in the square root is $4wv-u^2$