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A Romanian that one day is going to be like Ramanujan or far beyond.


1h
comment Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
@FelixMarin Thank you. :-)
2h
answered How find this sum closed form $I=\sum_{k=1}^{n}\int_{0}^{+\infty}\cos{(2kx)}x^{m-1}e^{-ax}dx$
18h
comment Proving that $\int_0 ^1 \frac{\text{d}s}{\sqrt{1-s^2}}$ converges with no trig functions
This is the best answer (+1)
1d
comment Alternative ways to evaluate $\displaystyle \int^1_0 \frac{\text{Li}_2(x)^3}{x}\,dx$
Thanks for your answer (+1)
1d
awarded  Favorite Question
1d
comment Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$
@robjohn yeah, right. I probably misinterpreted a limit.
1d
revised Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$
edited tags
1d
revised Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$
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1d
revised Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$
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1d
revised Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$
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2d
answered Proving $\int_{0}^{\infty}\frac{x}{(x^2+1)(e^{2\pi x}+1)} dx=1-\frac{\gamma}{2}-\ln2$
2d
answered Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$
Dec
15
comment If you have two envelopes, and …
The first important observation is that after letting $kx\mapsto x$ it's enough to consider the integral from $0$ to $1$. Having said that, we employ the simple fact that $-2\sum (-x)^k \cos(k a)/k=\log(x^2+2x \cos(a)+1)$. After that, we're immediately done.
Dec
14
comment An advanced integral $\int_0^1 \frac{(2 e)^{-1/y} \left(2 e^{1/y}-e 2^{1/y}\right)}{1-y} \ dy$
Very nicely done (+1)
Dec
14
accepted An advanced integral $\int_0^1 \frac{(2 e)^{-1/y} \left(2 e^{1/y}-e 2^{1/y}\right)}{1-y} \ dy$
Dec
14
revised An advanced integral $\int_0^1 \frac{(2 e)^{-1/y} \left(2 e^{1/y}-e 2^{1/y}\right)}{1-y} \ dy$
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Dec
14
asked An advanced integral $\int_0^1 \frac{(2 e)^{-1/y} \left(2 e^{1/y}-e 2^{1/y}\right)}{1-y} \ dy$
Dec
10
comment Evaluating the integral $\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$
@Lucian Yes, I went that way.
Dec
10
awarded  Nice Question
Dec
10
revised Evaluating the integral $\int_1^{\infty} \int_1^{\infty} \frac{\Gamma(x+1)\Gamma(y+1)}{x y \Gamma(x+y+2)} \ dx \ dy$
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