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

$\newcommand{\limitp}{\alpha}\newcommand{\innerp}{\beta}$I am fascinated by definite integrals. Exploring math.stackexchange, I have found many interesting integrals of the form $$\mathcal{J}(b,c,d;\innerp,\limitp) = \int_0^{\limitp} \, \frac{\log{(1+\innerp x^b)}}{(1+x^c)^d} \ \mathrm{d}x, \qquad \alpha,b,c,d\ge0,$$ the solutions to which have taught me a great deal. Many special cases have been evaluated, some with great difficulty and skill. Here is an attempt to collate and admire some of that work, and perhaps build upon it. Some of the techniques involved include series expansion, contour integration, differentiation with respect to $\innerp$ or maybe even $b$, and the answers often involve the evaluation of Euler sums and Hypergeometric functions, although sometimes it seems these difficulties can be bypassed.

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 $$ to . – Antonio Vargas Oct 28 '13 at 7:47 • Small nitpick with "we generate the conjecture". Mathematica evaluates the integral (after FullSimplify) to exactly that number on the RHS. Perhaps you are looking for a proof? – Alexander Vlasev Oct 30 '13 at 7:38 • Glad to meet a kindred spirit ! :-) I've also studied integrals of the following form recently:$$\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)$$– Lucian Oct 31 '13 at 9:03 • No proofs, I'm afraid. Just a curious, organized guy with a copy of Mathematica, lots of spare time, and a sense of observation. :-) I have a feeling that it's all based on expanding the Taylor series for \ln(1-t)=\sum_1^\infty\frac{t^n}n. – Lucian Oct 31 '13 at 23:43 • A more general formula would be$$\int_0^1\frac{\ln^nx\cdot\ln(1-x)}xdx=(-1)^{n-1}\cdot n!\cdot\zeta(n+2)$$– Lucian Nov 1 '13 at 0:09 1 Answer I know this is not an answer but I used Mathematica to obtain some expressions like the one you stated near the end of your post. The first few are for completeness. Here C is Catalan's constant. Let \mathcal{J}(k,2,2,1,\infty) = J_k. We have$$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. – Baby Dragon Oct 31 '13 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 $J_4$. – Bennett Gardiner Nov 4 '13 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. – Bennett Gardiner Nov 6 '13 at 12:03
• Yea, that sounds like a good idea – Alexander Vlasev Nov 6 '13 at 17:53