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Question Number 98967 by  M±th+et+s last updated on 17/Jun/20

$$solve:$$$$\left(\frac{{\int }_2^{6}x\sqrt{1+9\lfloor x{\rfloor }^2}dx}{{\int }_0^{1}x\left\{\frac{1}{x}\right\}\lceil \frac{1}{x}\rceil dx}\right)\left(\sum _{n⩾1}{(-1)}^n\frac{\prod _{j=1}^n(\frac{3}{2}-j)}{(2n+1)n!}\right)$$

Answered by maths mind last updated on 17/Jun/20

$$\sum _{n⩾1}{(-1)}^n.\frac{\underset{j=1}{\prod ^n}(\frac{3}{2}-j)}{(2n+1)n!}=\sum _{n⩾1}\frac{{(-1)}^n.\underset{j=0}{\prod ^{n-1}}(j-\frac{1}{2}).\underset{j=0}{\prod ^{n-1}}(j+\frac{1}{2})}{\underset{j=0}{\prod ^{n-1}}(\frac{3}{2}+j).n!}$$$${=}_2{F}_1(-\frac{1}{2},\frac{1}{2};\frac{3}{2};-1)=\frac{1}{\beta (\frac{1}{2},1)}{\int }_0^1{x}^{-\frac{1}{2}}.{(1-x)}^{\frac{3}{2}-\frac{1}{2}-1}{(1+x)}^{\frac{1}{2}}dx$$$$=\frac{\mathrm{\Gamma }\left(\frac{3}{2}\right)}{\mathrm{\Gamma }\left(\frac{1}{2}\right)\mathrm{\Gamma }\left(1\right)}.{\int }_0^1\sqrt{\frac{x+1}{x}}dx$$$$easynow$$$${\int }_0^{1}x\left\{\frac{1}{x}\right\}\left[\frac{1}{x}\right]dx={\int }_1^{+\mathrm{\infty }}\frac{\left\{y\right\}\left[y\right]}{y^3}dy$$$$=\sum _{n⩾1}{\int }_n^{n+1}\frac{(y-\left[y\right]]\left[y\right]}{y^3}dy$$$$=\sum _{n⩾1}{\int }_n^{n+1}(\frac{n}{y^2}-\frac{n^2}{y^3})dy$$$$=\sum _{n⩾1}[-\frac{n}{n+1}+1+\frac{1}{2}\frac{n^2}{{(n+1)}^2}-\frac{n^2}{2n^2}]$$$$=\sum _{n⩾1}\frac{(-2n(n+1)+{(n+1)}^2+n^2}{2{(n+1)}^2}=\frac{1}{2}\sum _{n⩾1}\frac{1}{{(n+1)}^2}=\frac{\zeta \left(2\right)-1}{2}=\frac{{\pi }^2}{12}-\frac{1}{2}$$$$just{\int }_2^{6}x\sqrt{1+9{\left[x\right]}^2}dxeasyone$$$$$$

Commented by  M±th+et+s last updated on 17/Jun/20

well done ����������

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