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Question Number 116361 by mohammad17 last updated on 03/Oct/20

Commented by mohammad17 last updated on 03/Oct/20

$$provethat$$

Answered by maths mind last updated on 05/Oct/20

$$\int \frac{e^{x}ln\left(x\right)}{x}dx$$$$bypart{u}^{\prime }=\frac{e^x}{x},v=ln\left(x\right),u=Ei\left(x\right),v^{\prime }=\frac{1}{x}$$$$\int \frac{e^{x}ln\left(x\right)}{x}dx=Ei\left(x\right)ln\left(x\right)-\int \frac{Ei\left(x\right)}{x}dx$$$$Ei\left(x\right)=\gamma +ln\left(x\right)+\sum _{k⩾1}\frac{x^k}{k.k!}$$$$\int \frac{e^{x}ln\left(x\right)}{x}dx=Ei\left(x\right)ln\left(x\right)-\int \frac{1}{x}(\gamma +ln\left(x\right)+\sum _{k⩾1}\frac{x^k}{k.k!})dx$$$$=Ei\left(x\right)ln\left(x\right)-\int \frac{\gamma }{x}dx-\int \frac{ln\left(x\right)}{x}dx-\sum _{k⩾1}\int \frac{x^{k-1}}{k.k!}dx$$$$=Ei\left(x\right)ln\left(x\right)-\gamma ln\left(x\right)-\frac{{ln}^2\left(x\right)}{2}-\sum _{k⩾0}\int \frac{x^k}{(k+1)(k+1)!}dx$$$$=Ei\left(x\right)ln\left(x\right)-\gamma ln\left(x\right)-\frac{{ln}^2\left(x\right)}{2}-\sum _{k⩾0}\frac{x^{k+1}}{{(k+1)}^2(k+1)!}+c$$$$\sum _{k⩾0}\frac{x^{k+1}}{{(k+1)}^2(k+1)!}=\sum _{k⩾0}\frac{k!.k!.k!}{{(k+1)}^2(k+1)!.k!.k!}.\frac{x^{k+1}}{k!}$$$$=x\sum _{k⩾0}\frac{k!.k!.k!}{(k+1)!(k+1)!(k+1)!}\frac{x^k}{k!}$$$$k!={\left(1\right)}_k,(k+1)!={\left(2\right)}_k$$$$=x\sum _{k⩾0}\frac{{\left(1\right)}_k{\left(1\right)}_k{\left(1\right)}_k}{{\left(2\right)}_k{\left(2\right)}_k{\left(2\right)}_k}.\frac{x^k}{k!}=x_3{F}_3(1,1,1;2,2,2;x)$$$$weGet$$$$\int \frac{e^{x}ln\left(x\right)}{x}dx=E_i\left(x\right)ln\left(x\right)-\gamma ln\left(x\right)-\frac{{ln}^2\left(x\right)}{2}-x{}_3{F}_3(1,1,1;2,2,2;x)+c$$$$=\frac{ln\left(x\right)}{2}(Ei\left(x\right)-2\gamma +ln\left(\frac{1}{x}\right))-x{}_3{F}_3(1,1,1;2,2,2;x)+c$$$$$$

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