# Further our knowledge of a certain class of integral involving logarithms.


My question then, is "can we build upon the work done in these answers, and develop a more general, complete theory for integrals of this type, or even extend our collection of specific cases?"

If you think you can evaluate a special case in a closed form, or even an interesting conjecture, answer here. Bounty goes to the best or most unique answer. I am particularly interested in larger values of $b,c$ and $d$, although feel free to make any contribution.

$\limitp=1$, $\innerp=1$, $b=1$, $c=2$, $d=1$

An attempt at a more general case,

$\limitp=1$, $\innerp=1$, $c=1$, $d=1$

A slightly less general case,

$\limitp=\infty$, $\innerp=1$, $b=4n$, $c=2$, $d=1$

Three integrals in one question, with contour integration featuring prominently,

$\limitp=\infty$, $\innerp=1$, $b=\left\{2,3,4\right\}$, $c=2$, $d=2$

Involves the golden ratio as a coefficient,

$\limitp=1$, $\innerp=\phi$, $b=2$, $c=1$, $d=2$

This one is truly amazing, involves an irrational exponent and fairly heavy number theoretic ideas,

$\limitp=1$, $\innerp=1$, $b=2+\sqrt{3}$, $c=1$, $d=1$

Related but slightly more general integrals, some with interesting solutions can be found here and here and here and here and here.

In particular, the case when $\limitp=\infty$, $\innerp=1$, $c=2$, $d=2$ seems interesting. Can we evaluate any cases for $b>4$? One case stands out as quite simple, when Mathematica evaluates the case $b=6$, we generate the result $$\mathcal{J}(6,2,2,1,\infty) = \frac{\pi}{4}\left(2\log 6 - 3\right).$$

• To get rid of the whitespace I changed the $$ to . Oct 28, 2013 at 7:47 • Small nitpick with "we generate the conjecture". Mathematica evaluates the integral (after FullSimplify) to exactly that number on the RHS. Perhaps you are looking for a proof? Oct 30, 2013 at 7:38 • Glad to meet a kindred spirit ! :-) I've also studied integrals of the following form recently:$$\int_0^1\frac{\ln^mx\cdot\ln^n(1-x)}{\qquad x^p\ \ \cdot\ \ (1-x)^q}dx$$all of which can be expressed as linear combinations of powers of the Riemann \zeta function, the most beautiful of which are:$$\int_0^1\frac{\ln(1-x)}xdx=-\zeta(2)\qquad\qquad\text{and}\qquad\qquad\int_0^1\frac{\ln x\cdot\ln(1-x)}xdx=\zeta(3)$$Oct 31, 2013 at 9:03 • No proofs, I'm afraid. Just a curious, organized guy with a copy of Mathematica, lots of spare time, and a sense of observation. :-) I have a feeling that it's all based on expanding the Taylor series for \ln(1-t)=\sum_1^\infty\frac{t^n}n. Oct 31, 2013 at 23:43 • A more general formula would be$$\int_0^1\frac{\ln^nx\cdot\ln(1-x)}xdx=(-1)^{n-1}\cdot n!\cdot\zeta(n+2)$$Nov 1, 2013 at 0:09 ## 1 Answer I know this is not an answer but I used Mathematica to obtain some expressions like the one you stated near the end of your post. The first few are for completeness. Here C is Catalan's constant. Let \mathcal{J}(k,2,2,1,\infty) = J_k. We have$$J_1 = \frac{C}{2}+\frac{\pi}{8} \log 2-\frac{\pi }{8}J_2 = \frac{\pi}{2} \log 2-\frac{\pi}{4}J_3 = -\frac{C}{6}+\frac{\pi}{24} \log \left(8 \left(2+\sqrt{3}\right)^8\right)-\frac{3 \pi }{8}J_4 = \frac{\pi}{4} \log \left(2(3+2 \sqrt{2})\right)-\frac{\pi }{2} J_5 = \frac{C}{10}-\frac{5 \pi }{8}-\frac{7\pi}{40} \log (2)+\frac{\pi}{5} \log \left(4+\sqrt{10-2 \sqrt{5}}\right)+\frac{\pi}{10} \log \left(43+7 \sqrt{5}+4 \sqrt{130+38 \sqrt{5}}\right)J_6 = \frac{\pi}{2} \log 6-\frac{3 \pi }{4}J_8 = \frac{\pi }{4} \log \left(34+16 \sqrt{2}+8 \sqrt{26+17 \sqrt{2}}\right)-\piJ_{10}= -\frac{5 \pi }{4}+\frac{\pi}{20} \log (80000000)-\frac{\pi}{5} \log \left(5-2 \sqrt{5}\right)+\frac{\pi}{20} \log \left(2889+1292 \sqrt{5}\right)$$I particularly like how Mathematica evaluates J_5 as an absolute horrifying mess. It includes a ton of square roots of various numbers and logarithms of 1\pm(-1)^{i/10}. But most of these can be simplified by manually replacing Log with the definition of the principal branch \log(z) = \ln|z|+i\arg(z) where \arg(z). The argument becomes an \arctan(x) which I replace with \arctan(FullSimplify[x]). The rest is just taking the real part, combining and cleaning up the roots and logs. FullSimplify has a lot of trouble at most stages and that's why I had to manually split a lot of the work. I'm currently trying to do J_7 in this manner. • Why do you suppose that J_8 is so much easier than J_9 or J_7? It looks like the odds have a term of the form \frac{C}{4n+2}. The \pi terms all seem to look like \frac{k\pi}{8}. Not quite a conjecture (perhaps a pre-conjecture) is that$$J_{2k+1}=(-1)^k\frac{C}{2k+4}-\frac{(2k+1)\pi}{8}+\mbox {some log terms}$$, while$$J_{2k}=\mbox{some log terms}-\frac{2k\pi}{8}.  Lastly I guessing that the prime factorization of the integer somehow will control the log terms. Oct 31, 2013 at 21:59
• Even though $J_6$ looks easier, $J_8$ might be easier to evaluate using contours, as the pole is also a branch point for the log in $J_6$, plus other branch points lie on the real line. In $J_8$ everything seems to be nice as in this evaluation of $J_4$. Nov 4, 2013 at 1:30
• I will award you the bounty for trying to get a discussion going, but leave the question open just in case someone else gets inspired. Nov 6, 2013 at 12:03
• Yea, that sounds like a good idea Nov 6, 2013 at 17:53