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Aug
13
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Hi Tunk. You may be onto something. Your thinking is kind of along the lines of mine. Using the incomplete beta function, I managed to derive an equivalent series for the left integral I said was tougher. It is $$1/64\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}\left(\psi_{2}(n/4+3/4)-\psi_{2}(n/4+‌​1/4)\right)$$. Now, if we choose the correct kernel, I think this can be done using residues as is done with the Euler sums. Maybe something like $$\pi\cdot csc(\pi z)\psi_{2}(z/4+3/4)$$. I have yet to try it. Random Variable has employed this method for various Euler sums across the site.
Aug
13
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
I merely posted it as a challenge problem thinking some may enjoy it. That is all. If not, please feel free to delete it. I have managed to solve a large part of it, but have not completed it entirely. I broke it up into $$\int_{0}^{1}\frac{\log^{2}(x)\log(1+x)}{x^{2}+1}dx+\int_{0}^{1}\frac{x\log^{3}‌​(x)}{(1-x)(x^{2}+1)}dx$$. The right integral is not too awful bad and can be done by using geometric series. It evaluates to $$-\frac{9\pi^{4}}{256}+\frac{1}{512}\left(\psi_{3}(1/4)-\psi_{3}(3/4)\right)$$. The other one is a little tougher, and I have not completed it yet.
Aug
12
asked Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Aug
10
awarded  Nice Question
Aug
10
awarded  definite-integrals
Aug
9
accepted Log trig integral with radical
Aug
9
revised log-trig integral with sin, cos, and tan
added 889 characters in body
Aug
9
revised log-trig integral with sin, cos, and tan
add derivation of sum
Aug
9
awarded  Nice Question
Aug
9
comment log-trig integral with sin, cos, and tan
:):):) Like me, I knew you'd would obsess until you got it. Cool.
Aug
8
revised log-trig integral with sin, cos, and tan
added 358 characters in body
Aug
8
answered log-trig integral with sin, cos, and tan
Aug
8
comment log-trig integral with sin, cos, and tan
I have seen that thread. That is a nice solution, M. But, this problem has $1+\sin(x), \;\ 1+\cos(x)$, which makes it more difficult. We can get rid of those by rewriting it as I mentioned in the above comment.
Aug
8
comment log-trig integral with sin, cos, and tan
Yes, M, I tried the Beta because it is often handy with these type of integrals. Alas, the solution may be presenting itself, but I am overlooking it. I am not so good at manipulating double sums like this. I thought perhaps if it can be done, some other clever soul to expand on it.
Aug
8
comment log-trig integral with sin, cos, and tan
If you feel up to it, I would very much like to see some of your work that led to this last line. I had arrived at an expression close to, but not as 'compact' as yours. By using Beta, I managed to arrive at a double Gamma sum: $$\displaystyle \sum_{k=1}^{\infty}\sum_{n=1}^{\infty}\frac{(-1)^{k+n}\Gamma(n/2+1)\Gamma(k/2)}{‌​nk\Gamma(n/2+k/2+1)}$$. This checks numerically, but I too am stuck here.
Aug
8
comment log-trig integral with sin, cos, and tan
I also tried that same sub, columbus. So, since there are some talented and brilliant mathematicians on SE, I thought perhaps they may like this one. I made a sub and arrived at the equivalent $\int_{0}^{1}\frac{\ln(1+\sqrt{1-x^{2}})\ln(x+1)}{x}dx$. I also tried writing it as $\int_{0}^{\frac{\pi}{2}}\frac{\ln(2\sin^{2}(\frac{x}{2}+\frac{\pi}{4}))\ln(2 \cos^{2}(x/2))\sin(x)}{\cos(x)}dx$, for what it's worth. :) I also derived a double sum involving Gamma, but it was certainly not any easier.
Aug
7
asked log-trig integral with sin, cos, and tan
Aug
7
revised Proof of $\int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\beta +x^{2}}=\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }$
added 25 characters in body
Aug
7
answered Proof of $\int_0^\infty \frac{x^{\alpha}dx}{1+2x\cos\beta +x^{2}}=\frac{\pi\sin (\alpha\beta)}{\sin (\alpha\pi)\sin \beta }$
Aug
7
revised Prove that $\int_0^1\frac{1-x}{1-x^6}\ln^4x\,dx=\frac{16\sqrt{3}}{729}\pi^5+\frac{605}{54}\zeta(5)$
fixed typo. should have been psi(2x) in first identity