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50s
revised Proving $\lim_{R \to \infty} \frac{1}{2 \pi i}\int_{\gamma_{R}} \frac{p(z)}{q(z)}\,dz = \frac{a_0}{b_0}$
deleted 5 characters in body
2m
revised Proving $\lim_{R \to \infty} \frac{1}{2 \pi i}\int_{\gamma_{R}} \frac{p(z)}{q(z)}\,dz = \frac{a_0}{b_0}$
added 13 characters in body; edited title
16m
answered Proving $\lim_{R \to \infty} \frac{1}{2 \pi i}\int_{\gamma_{R}} \frac{p(z)}{q(z)}\,dz = \frac{a_0}{b_0}$
59m
comment Proving $\lim_{R \to \infty} \frac{1}{2 \pi i}\int_{\gamma_{R}} \frac{p(z)}{q(z)}\,dz = \frac{a_0}{b_0}$
Are you familiar with the residue at infinity?
1d
revised hand evaluate $\sqrt{e}$
added 2 characters in body
1d
comment hand evaluate $\sqrt{e}$
I just realized that this is the series I derived in my answer but expressed in a slightly different way.
1d
revised hand evaluate $\sqrt{e}$
added 293 characters in body
1d
revised hand evaluate $\sqrt{e}$
deleted 7 characters in body
1d
answered hand evaluate $\sqrt{e}$
2d
revised $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$
added 163 characters in body
2d
revised $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$
added 156 characters in body
2d
revised $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$
added 79 characters in body
2d
revised $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$
edited tags
2d
revised $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$
added 79 characters in body
2d
answered $\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$
Apr
21
revised Compute $\int_0^\infty \frac{\ln x}{(1+x)^3}\,\mathrm{d}x$
added 2 characters in body
Apr
21
answered Compute $\int_0^\infty \frac{\ln x}{(1+x)^3}\,\mathrm{d}x$
Apr
20
comment Beautiful Closed form $\int_0^1 dx \frac{\ln x \ln^2(1-x)\ln(1+x)}{x}$
Sos440 gave an evaluation (which I helped simplify a bit) for the case where $\ln(1+x)$ is squared instead of $\ln(1-x)$. math.stackexchange.com/questions/560950/…
Apr
20
comment Beautiful Closed form $\int_0^1 dx \frac{\ln x \ln^2(1-x)\ln(1+x)}{x}$
Since the polylogs are being evaluated at $\frac{1}{2}$, I don't feel that the beauty of the answer is ruined.
Apr
20
revised Evaluating $\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)$
deleted 21 characters in body