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Question Number 138145 by mnjuly1970 last updated on 10/Apr/21

$$$$$$...........advanced.........calculus.........$$$$findthevalueof::$$$$\mathrm{\Theta }=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{\mathrm{H}}_{2n}}{n}=???$$

Answered by Dwaipayan Shikari last updated on 10/Apr/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_{2n}{x}^{2n}=H_2{x}^2+H_4{x}^4+H_6{x}^6+H_8{x}^8+...=\psi \left(x^2\right)$$$$And\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{H}_{2n}}{n}={\int }_0^1\frac{\psi (-x)}{\sqrt{x}}dx$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{H}_n{x}^n=H_{1}x+H_2{x}^2+H_3{x}^3+...=-\frac{log(1-x)}{1-x}$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}{(-1)}^n{H}_n{x}^n=-H_{1}x+H_2{x}^2-H_3{x}^3+...=-\frac{log(1+x)}{1+x}$$$$\psi \left(x^2\right)=-\frac{1}{2}(\frac{log(1-x)}{1-x}+\frac{log(1+x)}{1+x})$$$$\psi \left(x\right)=-\frac{1}{2}(\frac{log(1-\sqrt{x})}{1-\sqrt{x}}+\frac{log(1+\sqrt{x})}{1+\sqrt{x}})$$$${\int }_0^1\frac{\psi (-x)}{\sqrt{x}}=-\frac{1}{2}{\int }_0^1\frac{1}{i\sqrt{x}}(\frac{log(1-i\sqrt{x})}{1-i\sqrt{x}}+\frac{log(1+i\sqrt{x})}{1+i\sqrt{x}})dxx=t^2$$$$=-\frac{1}{2i}{\int }_0^1\frac{log(1-it)}{1-it}+\frac{log(1+it)}{1+it}dt$$$$=-\frac{1}{2i}{\int }_0^1\frac{log(1+t^2)}{1+t^2}+itlog\left(\frac{1-it}{1+it}\right)dt$$$$....$$$$$$

Commented by mnjuly1970 last updated on 10/Apr/21

$$gratefulforyourattention$$$$andeffort...$$

Answered by Ñï= last updated on 10/Apr/21

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{n}{H}_{2n}=-{\int }_0^1\frac{1}{1-u}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}{x}^n(1-u^{2n})dux=1$$$$={\int }_0^1\frac{1}{u-1}\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{n-1}}{n}[x^n-{\left({xu}^2\right)}^n]du$$$$={\int }_0^1\frac{1}{u-1}[ln(1+x)-ln(1+{xu}^2)]du$$$$=.....???$$

Commented by mnjuly1970 last updated on 10/Apr/21

$$thanksalotmr...$$

Answered by mnjuly1970 last updated on 10/Apr/21

$$f\left(x\right):=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_n}{n}{x}^n=\frac{1}{2}{ln}^2(1-x)+{li}_2\left(x\right)$$$$f\left(i\right):=\mathrm{\Sigma }\frac{{\mathrm{H}}_n}{n}{\left(i\right)}^n=\frac{1}{2}{ln}^2(1-i)+{li}_2\left(i\right)....(\ast )$$$$f(-i):=\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_n}{n}{(-i)}^n=\frac{1}{2}{ln}^2(1+i)+{li}_2(-i)....(\ast \ast )$$$$(\ast )+(\ast \ast ):\mathrm{\Sigma }\frac{{\mathrm{H}}_n}{n}(i^n+{(-i)}^n)=\frac{1}{2}\{{ln}^2\left(\sqrt{2}{e}^{\frac{-i\pi }{4}}\right)+{ln}^2\left(\sqrt{2}{e}^{\frac{i\pi }{4}}\right)\}+\frac{1}{2}{li}_2\left(i^2\right)$$$$\therefore \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{{\mathrm{H}}_{2n}}{2n}({(-1)}^n+{(-1)}^n)=\frac{1}{2}\{{(ln\left(\sqrt{2}\right)-\frac{i\pi }{4})}^2+{(ln\left(\sqrt{2}\right)+\frac{i\pi }{4})}^2\}-\frac{{\pi }^2}{24}$$$$=\frac{1}{2}\left\{2(\frac{1}{4}{ln}^2\left(2\right)-\frac{{\pi }^2}{16})\right\}-\frac{{\pi }^2}{24}$$$$\therefore \mathrm{\Theta }=\frac{1}{4}{ln}^2\left(2\right)-\frac{5{\pi }^2}{48}...✓✓$$

Answered by Kamel last updated on 10/Apr/21

Commented by Kamel last updated on 10/Apr/21

$$Withoutcomplexnumbers.$$

Commented by mnjuly1970 last updated on 10/Apr/21

$$thanksalot..$$$$$$

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