Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 137011 by mohammad17 last updated on 28/Mar/21

Commented by mohammad17 last updated on 28/Mar/21

$$helpmesirplease$$

Commented by mohammad17 last updated on 28/Mar/21

$$?????$$

Answered by Dwaipayan Shikari last updated on 28/Mar/21

$${\int }_0^1\frac{\sqrt{y}}{1+\sqrt[5]{y}}dyy=t^5$$$$5{\int }_0^1\frac{t^{\frac{13}{2}}}{1+t}dt=5{\int }_0^1\frac{t^{\frac{13}{2}}-t^{\frac{15}{2}}}{1-t^2}dt{t}^2=u$$$$=\frac{5}{2}{\int }_0^1\frac{u^{\frac{11}{4}}-u^{\frac{13}{4}}}{1-u}du=\frac{5}{2}(\psi \left(\frac{17}{4}\right)-\psi \left(\frac{11}{4}\right))$$$$=\frac{5}{2}(\psi \left(\frac{13}{4}\right)+\frac{4}{13}-\psi \left(\frac{7}{4}\right)-\frac{4}{7})=\frac{5}{2}(\psi \left(\frac{9}{4}\right)+\frac{4}{13}+\frac{4}{9}-\frac{4}{7}-\frac{4}{3}-\psi \left(\frac{3}{4}\right))$$$$=\frac{5}{2}(\psi \left(\frac{1}{4}\right)-\psi \left(\frac{3}{4}\right))+10(1+\frac{1}{5}+\frac{1}{9}+\frac{1}{13}-\frac{1}{7}-\frac{1}{3})$$$$=-\frac{5\pi }{2}+10(\frac{59}{45}-\frac{10}{21}+\frac{1}{13})=10(\frac{263}{315}+\frac{1}{13})-\frac{5\pi }{2}=\frac{526}{63}+\frac{10}{13}-\frac{5\pi }{2}$$

Commented by mohammad17 last updated on 28/Mar/21

$$sirwhatsthevalueof\psi$$

Commented by Dwaipayan Shikari last updated on 28/Mar/21

$$Generally$$$$\psi (1-s)-\psi \left(s\right)=\pi cot\left(\pi s\right)$$$$And\psi (1+s)=\psi \left(s\right)+\frac{1}{s}\left(Ihavedonethisreduction\right)$$

Answered by mathmax by abdo last updated on 29/Mar/21

$$1)\mathrm{I}={\int }_0^1\frac{\sqrt{\mathrm{x}}}{1+{\mathrm{x}}^{\frac{1}{5}}}\mathrm{dx}\left(\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}\right)\Rightarrow \mathrm{I}{=}_{\sqrt{\mathrm{x}}=\mathrm{t}}{\int }_0^1\frac{\mathrm{t}}{1+{\mathrm{t}}^{\frac{2}{5}}}\left(2\mathrm{t}\right)\mathrm{dt}$$$$=2{\int }_0^1\frac{{\mathrm{t}}^2}{1+{\mathrm{t}}^{\frac{2}{5}}}\mathrm{dt}\mathrm{changement}{\mathrm{t}}^{\frac{2}{5}}=\mathrm{y}\mathrm{give}\mathrm{t}={\mathrm{y}}^{\frac{5}{2}}\Rightarrow$$$$\mathrm{I}=2.\frac{5}{2}{\int }_0^1\frac{{\mathrm{y}}^5}{1+\mathrm{y}}{\mathrm{y}}^{\frac{5}{2}-1}\mathrm{dy}=5{\int }_0^1\frac{{\mathrm{y}}^{\frac{15}{2}-1}}{1+\mathrm{y}}\mathrm{dy}$$$$=5{\int }_0^1{\mathrm{y}}^{\frac{13}{2}}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\mathrm{y}}^{\mathrm{n}}\mathrm{dy}$$$$=5\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}{\int }_0^1{\mathrm{y}}^{\mathrm{n}+\frac{13}{2}}\mathrm{dy}$$$$=5\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{(-1)}^{\mathrm{n}}\frac{1}{\mathrm{n}+\frac{13}{2}+1}{\left[{\mathrm{y}}^{\mathrm{n}+\frac{13}{2}+1}\right]}_0^1$$$$=5\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{\mathrm{n}+\frac{15}{2}}=10\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{2\mathrm{n}+15}$$$$\mathrm{rest}\mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{serie}...\mathrm{be}\mathrm{continued}...$$

Answered by mathmax by abdo last updated on 29/Mar/21

$$\mathrm{\Phi }={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-({\mathrm{y}}^2-3\mathrm{y})}{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{dy}\Rightarrow \mathrm{\Phi }={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\{{\mathrm{y}}^2-2\frac{3}{2}\mathrm{y}+\frac{9}{4}-\frac{9}{4}\}\times \frac{1}{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{dy}$$$$={\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{\frac{-{(\mathrm{y}-\frac{3}{2})}^2+\frac{9}{4}}{{\mathrm{a}}^2+{\mathrm{b}}^2}}\mathrm{dy}={\mathrm{e}}^{\frac{9}{4({\mathrm{a}}^2+{\mathrm{b}}^2)}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-{\left(\frac{\mathrm{y}-\frac{3}{2}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}\right)}^2}\mathrm{dy}$$$${=}_{\frac{\mathrm{y}-\frac{3}{2}}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}=\mathrm{z}}{\mathrm{e}}^{\frac{9}{4({\mathrm{a}}^2+{\mathrm{b}}^2)}}.{\int }_{-\frac{3}{2\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}^{\mathrm{\infty }}{\mathrm{e}}^{-{\mathrm{z}}^2}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\mathrm{dz}$$$$=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}.{\mathrm{e}}^{\frac{9}{4({\mathrm{a}}^2+{\mathrm{b}}^2)}}\{{\int }_{-\frac{3}{2\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}^0{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}+\frac{\sqrt{\pi }}{2}\}$$$$\mathrm{we}\mathrm{have}{\int }_{-\frac{3}{2\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}^0{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}{=}_{\mathrm{z}=-\mathrm{u}}-{\int }_0^{\frac{3}{2\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}{\mathrm{e}}^{-{\mathrm{u}}^2}(-\mathrm{du})\Rightarrow$$$$\mathrm{\Phi }=\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}.{\mathrm{e}}^{\frac{9}{4({\mathrm{a}}^2+{\mathrm{b}}^2)}}\{\frac{\sqrt{\pi }}{2}+{\int }_0^{\frac{3}{2\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}}{\mathrm{e}}^{-{\mathrm{z}}^2}\mathrm{dz}\}$$

Answered by mathmax by abdo last updated on 29/Mar/21

$$3)\mathrm{I}={\int }_{-1}^1\frac{{\mathrm{e}}^{2\mathrm{y}}}{\sqrt{7}+{\mathrm{e}}^{\mathrm{y}}}\mathrm{dy}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{chanhement}{\mathrm{e}}^{\mathrm{y}}=\mathrm{x}\Rightarrow$$$$\mathrm{I}={\int }_{\frac{1}{\mathrm{e}}}^{\mathrm{e}}\frac{{\mathrm{x}}^2}{\sqrt{7}+\mathrm{x}}\frac{\mathrm{dx}}{\mathrm{x}}={\int }_{\frac{1}{\mathrm{e}}}^{\mathrm{e}}\frac{\mathrm{x}}{\mathrm{x}+\sqrt{7}}\mathrm{dx}={\int }_{\frac{1}{\mathrm{e}}}^{\mathrm{e}}\frac{\mathrm{x}+\sqrt{7}-\sqrt{7}}{\mathrm{x}+\sqrt{7}}\mathrm{dx}$$$$={\left[\mathrm{x}\right]}_{\frac{1}{\mathrm{e}}}^{\mathrm{e}}-\sqrt{7}{\left[\ln (\mathrm{x}+\sqrt{7})\right]}_{\frac{1}{\mathrm{e}}}^{\mathrm{e}}=\mathrm{e}-{\mathrm{e}}^{-1}-\sqrt{7}\{\ln (\mathrm{e}+\sqrt{7})-\ln ({\mathrm{e}}^{-1}+\sqrt{7})\}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com