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Question Number 74014 by mathmax by abdo last updated on 17/Nov/19

$$$$ $$letW\left(x\right)=\sum _{1⩽i<j⩽n}\frac{x^{i+j}}{ij}$$ $$calculateW{}^{{}^{\prime }}\left(x\right).$$

Commented bymathmax by abdo last updated on 19/Nov/19

$$wehave{\left(\sum _{i=1}^n\frac{x^i}{i}\right)}^2=\sum _{i=1}^n\frac{x^{2i}}{i^2}+\sum _{1⩽i<j⩽n}\frac{x^{i+j}}{ij}\Rightarrow$$ $$W\left(x\right)={\left(\sum _{i=1}^n\frac{x^i}{i}\right)}^2-\sum _{i=1}^n\frac{x^{2i}}{i^2}\Rightarrow$$ $$W{}^{{}^{\prime }}\left(x\right)=2{\left(\sum _{i=1}^n\frac{x^i}{i}\right)}^{{}^{\prime }}\left(\sum _{i=1}^n\frac{x^i}{i}\right)-\sum _{i=1}^n\frac{2i{x}^{2i-1}}{i^2}$$ $$=2\left(\sum _{i=1}^n{x}^{i-1}\right)\left(\sum _{i=1}^n\frac{x^i}{i}\right)-2\sum _{i=1}^n\frac{x^{2i-1}}{i}wehave$$ $$\sum _{i=1}^n{x}^{i-1}=\sum _{i=0}^{n-1}{x}^i=\frac{1-x^n}{1-x}ifx\ne 1and\sum _{i=1}^n{x}^{i-1}=nifx=1$$ $$letf\left(x\right)=\sum _{i=1}^n\frac{x^i}{i}\Rightarrow f^{{}^{\prime }}\left(x\right)=\sum _{i=1}^n{x}^{i-1}=\frac{1-x^n}{1-x}(ifx\ne 1)$$ $$f\left(x\right)={\int }_0^x\frac{1-t^n}{1-t}dt+c(c=f\left(0\right)=0)\Rightarrow f\left(x\right)={\int }_0^x\frac{1-t^n}{1-t}dt$$ $$={\int }_0^x\frac{dt}{1-t}-{\int }_0^x\frac{t^n}{1-t}dt=-ln\mid 1-x\mid -{\int }_0^x\frac{t^n}{1-t}dt$$ $$x=1\Rightarrow f\left(x\right)=\sum _{i=1}^n\frac{1}{i}=H_{n}also$$ $$\sum _{i=1}^n\frac{x^{2i-1}}{i}=\frac{1}{x}\sum _{i=1}^n\frac{{\left(x^2\right)}^i}{i}=\frac{1}{x}f\left(x^2\right)=\frac{1}{x}(-ln\mid 1-x^2\mid -{\int }_0^{x^2}\frac{t^n}{1-t}dt)$$ $$=-\frac{1}{x}(ln\mid 1-x^2\mid +{\int }_0^{x^2}\frac{t^n}{1-t}dt)(x\ne 1)soforx\ne 1$$ $$W^{{}^{\prime }}\left(x\right)=2\left(\frac{x^n-1}{x-1}\right)(ln\mid 1-x\mid +{\int }_0^x\frac{t^n}{1-t}dt)+\frac{2}{x}(ln\mid 1-x^2\mid +{\int }_0^{x^2}\frac{t^n}{1-t}dt)$$ $$$$ $$$$

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