Question Number 143045 by 0731619 last updated on 09/Jun/21
Answered by Dwaipayan Shikari last updated on 09/Jun/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{n}} }{\mathrm{1}−{x}}{dx}=−\gamma+\psi\left({n}+\mathrm{1}\right) \\ $$$$\underset{{n}\rightarrow\mathrm{3}} {\mathrm{lim}}\frac{−\gamma−\mathrm{1}.\mathrm{833}+\psi\left({n}+\mathrm{1}\right)}{\Gamma\left({n}+\mathrm{1}\right)−\mathrm{6}}\overset{{Lhopital}} {=}\frac{\psi'\left({n}+\mathrm{1}\right)}{\Gamma'\left({n}+\mathrm{1}\right)}=\frac{\psi'\left(\mathrm{4}\right)}{\psi\left(\mathrm{4}\right)\Gamma\left(\mathrm{4}\right)} \\ $$$$\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}+\mathrm{4}\right)^{\mathrm{2}} }}{\left(\gamma+\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}}\right)\mathrm{6}}=\frac{\frac{\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }−\mathrm{1}}{\mathrm{6}\gamma+\frac{\mathrm{25}}{\mathrm{2}}}=\frac{\mathrm{6}\pi^{\mathrm{2}} −\mathrm{294}}{\mathrm{216}\gamma+\mathrm{450}} \\ $$
Commented by Canebulok last updated on 11/Jun/21
Hello mr. Payan , are you a mathematician?
Commented by Dwaipayan Shikari last updated on 11/Jun/21
Thanks for asking. I am a student and I am trying to learn my interest