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Question Number 64677 by mathmax by abdo last updated on 20/Jul/19

$$letf\left(x\right)={\int }_0^{1}lntln(1-xt)dtwith\mid x\mid <1$$$$1)determineaexplicitformforf\left(x\right)$$$$2)findalsog\left(x\right)={\int }_0^1\frac{tlnt}{1-xt}dt$$$$3)give{f}^{\left(n\right)}\left(x\right)atformofintegral$$$$4)calculate{\int }_0^{1}ln\left(t\right)ln(1-t)dtand{\int }_0^{1}ln\left(t\right)ln(2-t)dt$$$$5)calculate{\int }_0^1\frac{tln\left(t\right)}{2-t}dt.$$

Commented by mathmax by abdo last updated on 21/Jul/19

$$1)f\left(x\right)={\int }_0^{1}lntln(1-xt)dtcase10<x<1changementxt=u$$$$givef\left(x\right)={\int }_0^{x}ln\left(\frac{u}{x}\right)ln(1-u)\frac{du}{x}=\frac{1}{x}{\int }_0^x(lnu-lnx)ln(1-u)du$$$$=\frac{1}{x}{\int }_0^{x}ln\left(u\right)ln(1-u)du-\frac{lnx}{x}{\int }_0^{x}ln(1-u)du$$$${\int }_0^{x}ln(1-u)du{=}_{1-u=z}{\int }_1^{1-x}ln\left(z\right)(-dz)={\int }_{1-x}^{1}ln\left(z\right)dz$$$${[zlnz-z]}_{1-x}^1=-1-((1-x)ln(1-x)-(1-x))$$$$=-1-(1-x)ln(1-x)+1-x=-x-(1-x)ln(1-x)$$$$alsowehave{ln}^{{}^{\prime }}(1-u)=\frac{-1}{1-u}=-\sum _{n=0}^{\mathrm{\infty }}{u}^n\Rightarrow$$$$ln(1-u)=-\sum _{n=0}^{\mathrm{\infty }}\frac{u^{n+1}}{n+1}+c(c=0)=-\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}\Rightarrow$$$${\int }_0^{x}ln\left(u\right)ln(1-u)du=-{\int }_0^{x}ln\left(u\right)\sum _{n=1}^{\mathrm{\infty }}\frac{u^n}{n}du$$$$=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}{\int }_0^x{u}^{n}lnudubyparts{f}^{{}^{\prime }}=u^{n}andg=lnu\Rightarrow$$$$w_n={\int }_0^x{u}^{n}ln\left(u\right)du={\left[\frac{1}{n+1}{u}^{n+1}lnu\right]}_0^x-{\int }_0^x\frac{u^{n+1}}{n+1}\frac{du}{u}$$$$=\frac{lnx}{n+1}{x}^{n+1}-\frac{1}{n+1}{\int }_0^x{u}^{n}du=\frac{lnx}{n+1}{x}^{n+1}-\frac{1}{n+1}{\left[\frac{u^{n+1}}{n+1}\right]}_0^x$$$$=\frac{lnx}{n+1}{x}^{n+1}-\frac{x^{n+1}}{{(n+1)}^2}\Rightarrow$$$${\int }_0^{x}ln\left(u\right)ln(1-u)du=-\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}\{\frac{lnx}{n+1}{x}^{n+1}-\frac{x^{n+1}}{{(n+1)}^2}\}$$$$=-lnx\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+1}}{n(n+1)}+\sum _{n=1}^{\mathrm{\infty }}\frac{x^{n+1}}{n{(n+1)}^2}\Rightarrow$$$$f\left(x\right)=-lnx\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n(n+1)}+\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{(n+1)}^2}-\frac{lnx}{x}(-x-(1-x)ln(1-x))$$$$f\left(x\right)=-lnx\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n(n+1)}+\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{(n+1)}^2}+lnx(1+(1-x)ln(1-x)).$$$$$$

Commented by mathmax by abdo last updated on 21/Jul/19

$$wehave\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n(n+1)}=\sum _{n=1}^{\mathrm{\infty }}(\frac{1}{n}-\frac{1}{n+1})x^n$$$$=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}=-ln(1-x)-\sum _{n=2}^{\mathrm{\infty }}\frac{x^{n-1}}{n}$$$$=-ln(1-x)-\frac{1}{x}(\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-x)$$$$=-ln(1-x)+\frac{1}{x}ln(1-x)+1=(\frac{1}{x}-1)ln(1-x)+1let$$$$decomposeF\left(t\right)=\frac{1}{t{(t+1)}^2}\Rightarrow F\left(t\right)=\frac{a}{t}+\frac{b}{t+1}+\frac{c}{{(t+1)}^2}$$$$a={lim}_{t\to 0}tF\left(t\right)=1$$$$c={lim}_{t\to -1}{(t+1)}^{2}F\left(t\right)=-1\Rightarrow F\left(t\right)=\frac{1}{t}+\frac{b}{t+1}-\frac{1}{{(t+1)}^2}$$$${lim}_{t\to +\mathrm{\infty }}tF\left(t\right)=0=a+b\Rightarrow b=-1\Rightarrow F\left(t\right)=\frac{1}{t}-\frac{1}{t+1}-\frac{1}{{(t+1)}^2}\Rightarrow$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{(n+1)}^2}=\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n+1}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{{(n+1)}^2}$$$$=-ln(1-x)-\sum _{n=2}^{\mathrm{\infty }}\frac{x^{n-1}}{n}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{{(n+1)}^2}$$$$=-ln(1-x)-\frac{1}{x}\{\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n}-1\}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{{(n+1)}^2}$$$$=-ln(1-x)+\frac{ln(1-x)}{x}+\frac{1}{x}-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{{(n+1)}^2}$$

Commented by mathmax by abdo last updated on 21/Jul/19

$$\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{n{(n+1)}^2}=-ln(1-x)+\frac{ln(1-x)}{x}+1-\sum _{n=1}^{\mathrm{\infty }}\frac{x^n}{{(n+1)}^2}\Rightarrow$$$$f\left(x\right)=$$

Commented by mathmax by abdo last updated on 22/Jul/19

$$2)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{-t}{1-xt}lntdt=-{\int }_0^1\frac{tlnt}{1-xt}dt=-g\left(x\right)\Rightarrow$$$$g\left(x\right)=-f^{{}^{\prime }}\left(x\right)thefunctionfisknownresttocalculate{f}^{{}^{\prime }}\left(x\right)$$$$3)wehave{f}^{{}^{\prime }}\left(x\right)={\int }_0^1\frac{tlnt}{xt-1}dt={\int }_0^1\frac{lnt}{x-\frac{1}{t}}dt\Rightarrow$$$$f^{\left(n\right)}\left(x\right)={\int }_0^1{\left(\frac{1}{x-\frac{1}{t}}\right)}^{(n-1)}lntdt={\int }_0^1\frac{{(-1)}^{n-1}(n-1)!}{{(x-\frac{1}{t})}^n}lntdt$$$$f^{\left(n\right)}\left(x\right)={(-1)}^{n-1}(n-1)!{\int }_0^1\frac{t^{n}lnt}{{(xt-1)}^n}dt$$

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