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While I was working on computing $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$, I came across the integral:

$$I=\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}\mathrm{d}x.$$

I tried $x=\sin(u)$ then $\tan(u/2)=t$ and got

$$I=\int_0^{\pi/2}\cot(u)\ln(\sin u)\ln(1-\sin u)\ln(1+\cos u)\mathrm{d}u$$

$$=\int_0^1\frac{1-t^2}{t(1+t^2)}\ln\left(\frac{2t}{1+t^2}\right)\ln\left(\frac{(1-t)^2}{1+t^2}\right)\ln\left(\frac{2}{1+t^2}\right)\mathrm{d}t.$$

Any idea?

Thanks

Ali Shadhar
  • 25,498
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    Does this answer your question? Closed form expression for the harmonic sum $\sum\limits_{n=1}^{\infty}\frac{H_{2n}}{n^2\cdot4^n}{2n \choose n}$ – VIVID Jun 18 '21 at 20:51
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    @VIVID The sum you've mentioned is different in the sense that the OP is asking for the generalized harmonic numbers where as the presented sum by you is for harmonic numbers only. – Amrit Awasthi Jun 18 '21 at 21:11
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    For answer-checking, $$I = -2 \pi \Im \text{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)-\frac{\text{Li}_4\left(\frac{1}{2}\right)}{2}+\frac{7}{16} \zeta (3) \log (2)+\frac{281 \pi ^4}{5760}-\frac{1}{48} \log ^4(2)-\frac{5}{48} \pi ^2 \log ^2(2)$$ – pisco Jun 18 '21 at 23:18
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    A generalization: $$\int_0^1\frac{\ln^2(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x} dx $$ equals $$24 \pi \Im\text{Li}_4(\frac{1}{2}+\frac{i}{2})+4 \pi \log (2) \Im\text{Li}_3 (\frac{1}{2}+\frac{i}{2})-20 \pi \beta(4)-\text{Li}_5\left(\frac{1}{2}\right)-\text{Li}_4\left(\frac{1}{2}\right) \log (2)-\frac{43 \pi ^2 \zeta (3)}{192}+\frac{1047 \zeta (5)}{128}+\frac{7}{16} \zeta (3) \log ^2(2)-\frac{1}{30} \log ^5(2)-\frac{7}{72} \pi ^2 \log ^3(2)+\frac{5}{32} \pi ^4 \log (2)$$ – pisco Jun 18 '21 at 23:26
  • @pisco thank you. – Ali Shadhar Jun 18 '21 at 23:53
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    @pisco do you have such generalization in some article? If so, what page? – Ali Shadhar Jun 18 '21 at 23:58
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    @AliShadhar There is an old proverb saying "give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime." For your sum or integrals, there are hundred of thousands, if not millions, variations one can write down. Rather than writing down and solving each of these hundred of thousands problems one by one, I chose to develop a systematic approach: if you give me any one anytime, I can perform this systematic procedure, and obtain the result. There is no article or book that I copied the above results from, they are produced from the systematic procedure. – pisco Jun 19 '21 at 00:20
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    @pisco congrats on designing such powerful systematic techniques. – Ali Shadhar Jun 19 '21 at 01:07

2 Answers2

18

Since you are primarily interested in that series I'll evaluate it. $$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$ Consider the well-known expansion: $$2\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}x^{2k}=\frac{\arcsin ^2\left(x\right)}{\sqrt{1-x^2}}$$ $$-4\sum _{k=1}^{\infty }\frac{\left(H_{2k}^{\left(2\right)}-\frac{1}{4}H_k^{\left(2\right)}\right)}{4^k}\binom{2k}{k}\int _0^1x^{2k-1}\ln \left(x\right)\:dx=-2\int _0^1\frac{\ln \left(x\right)\arcsin ^2\left(x\right)}{x\sqrt{1-x^2}}\:dx$$ $$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-2\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\ln \left(\sin \left(x\right)\right)\:dx+\frac{1}{4}\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$ You can find that integral here. I can also provide a new way to calculate it but it's lenghty.

For the remaining sum consider: $$\int _0^1x^{k-1}\ln ^2\left(1-x\right)\:dx=\frac{H_k^2+H_k^{\left(2\right)}}{k}$$ $$\int _0^1\left(\sum _{k=1}^{\infty }\frac{1}{4^kk}\binom{2k}{k}x^k\right)\frac{\ln ^2\left(1-x\right)}{x}\:dx=\sum _{k=1}^{\infty }\frac{H_k^2}{4^kk^2}\binom{2k}{k}+\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}$$ $$\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-16\underbrace{\int _0^1\frac{x\ln ^2\left(x\right)\ln \left(1+x\right)}{1-x^2}\:dx}_{I}+2\ln \left(2\right)\int _0^1\frac{\ln ^2\left(1-x\right)}{x}\:dx-\underbrace{\sum _{k=1}^{\infty }\frac{H_k^2}{4^kk^2}\binom{2k}{k}}_{S}$$ $$\sum _{k=1}^{\infty }\frac{H_k^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=3\zeta \left(4\right)-3\ln \left(2\right)\zeta \left(3\right)$$ $I$ transforms into known integrals after integration by parts and $S$ can be calculated by simple means as shown in the OP's book, page $\#297$

And therefore: $$\sum _{k=1}^{\infty }\frac{H_{2k}^{\left(2\right)}}{4^kk^2}\binom{2k}{k}=-\frac{129}{8}\zeta \left(4\right)+\frac{25}{4}\ln \left(2\right)\zeta \left(3\right)-\frac{3}{2}\ln ^2\left(2\right)\zeta \left(2\right)+8\pi \operatorname{\mathfrak{I}} \left\{\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right\}$$

Jorge Layja
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Bonus: Differentiate both sides of $\int_0^1 x^{2n-1}\ln(1-x)dx=-\frac{H_{2n}}{2n}$, we get

$$\int_0^1 x^{2n-1}\ln(x)\ln(1-x)dx=\frac{H_{2n}}{4n^2}+\frac{H_{2n}^{(2)}}{2n}-\frac{\zeta(2)}{2n}.$$

Multiply both sides by $\frac{2n\choose n}{4^nn}$ then sum up from $n=1$ to $\infty$, we obtain

$$\frac14\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}}{4^n n^3}+\frac12\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}-\zeta(2)\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n n^2}=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{{2n\choose n}x^{2n}}{4^n n}\right)dx$$

$$=\int_0^1\frac{\ln(x)\ln(1-x)}{x}\left(2\ln\left(\frac{2}{1+\sqrt{1-x^2}}\right)\right)dx$$

$$=2\ln(2)\int_0^1\frac{\ln(x)\ln(1-x)}{x}dx-2\int_0^1\frac{\ln(x)\ln(1-x)\ln(1+\sqrt{1-x^2})}{x}dx$$ $$=2\ln(2)\zeta(3)-2I.$$

The sum $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}^{(2)}}{4^n n^2}$ is calculated by Jorge above and the sum $\sum_{n=1}^\infty\frac{{2n\choose n}H_{2n}}{4^n n^3}$ is calculated here. Collecting these two results along with using $\sum_{n=1}^\infty\frac{{2n\choose n}}{4^n n^2}=\zeta(2)-2\ln^2(2)$, we get

$$I = -2 \pi \Im \left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}-\frac12\text{Li}_4\left(\frac{1}{2}\right)+\frac{281 }{64}\zeta(4)+\frac{7}{16} \ln (2)\zeta (3)-\frac{5}{8} \ln ^2(2)\zeta(2)-\frac{1}{48} \ln ^4(2).$$

Ali Shadhar
  • 25,498