Cody
Reputation
6,393
Top tag
Next privilege 10,000 Rep.
Access moderator tools
 Aug 23 comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$ Thanks LT. I just noticed your comment. I have not been around much lately due to work and other matters. Thanks. If I ever find a way to evaluate that tetragamma sum, I will surely post it ASAP. I thought perhaps some other learned individual may come up with something, but it appears to be pretty tough. Coffey wrote a paper on these, but using his methods it just loops back around to the original integral. I still think residues may be an option, but approaching it correctly is the trick. Aug 18 awarded Nice Question Aug 16 comment Closed form for the integral $\int_{0}^{\infty}\frac{\ln^{2}(x)\ln(1+x)}{(1-x)(x^{2}+1)}dx$ I played with this a little and managed to get the terms in the solution except for the log term. I can not determine how it comes in. I tried $\displaystyle \pi csc(\pi z)\left[\psi_{2}(z/4+3/4)-\psi_{2}(z/4+1/4)\right]$. 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.