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  • 176 votes cast
Aug
16
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Please work it out?. If I could do that I would have already posted the method. As I previously stated, I suspect the Flajolet/Salvy contour method may work, but finding the correct kernel, if it exists, proves tricky. Since digamma and its derivatives are directly related to Euler sums, this method can be used to evaluate digamma series as well. i.e. $\displaystyle \sum_{n=0}^{\infty}\frac{(-1)^{n}\psi_{2}(n+1)}{2n+1}$ can be found by using the kernel $\displaystyle \pi csc(\pi z)\psi_{2}(-z)$
Aug
15
answered Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Aug
15
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Vladimir posted it on I&S. I thought it was a nice one so I thought I would share. Hence, I do not understand the two down votes. I have a few more I can post if others are interested. I managed to get the left integral to a series involving tetragamma that I mention above, but evaluating that series has proved the issue. The right one is easier and not a problem.
Aug
13
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Good work xpaul. What you done at the top was exactly how I approached it to arrive at that alternating tetragamma series I mentioned. But, I am stuck there for now. Tunk's contour idea is also consideration.
Aug
13
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Yes, good idea.It seems to me RonG may have done something similar.
Aug
13
revised Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
fixed negative sign
Aug
13
comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$
Tunk, your idea looks like it may be a good one. Since $\alpha$ and $\beta$ are ultimately 0 and 1 and not fractional, then residues may not be too horrible. Looks like a good job for some of our ingenious contour pros like RonG, robjohn, RV, achille, etc. :)
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.