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Question Number 115920 by mnjuly1970 last updated on 29/Sep/20

$$$$$$provethat::$$$$$$$${\int }_0^{\mathrm{\infty }}({tanh}^a\left(x\right)-{tanh}^b\left(x\right))dx$$$$\stackrel{???}{=}\frac{\psi \left(\frac{b+1}{2}\right)-\psi \left(\frac{a+1}{2}\right)}{2}$$$$m.n.july.1970$$$$$$

Answered by mathmax by abdo last updated on 29/Sep/20

$$\mathrm{I}=\phi \left(\mathrm{a}\right)-\phi \left(\mathrm{b}\right)\mathrm{with}\phi \left(\mathrm{a}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{th}}^{\mathrm{a}}\left(\mathrm{x}\right)\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\left(\frac{\mathrm{shx}}{\mathrm{chx}}\right)}^{\mathrm{a}}\mathrm{dx}={\int }_0^{\mathrm{\infty }}{\left(\frac{{\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{-\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}+{\mathrm{e}}^{-\mathrm{x}}}\right)}^{\mathrm{a}}\mathrm{dx}$$$$={\int }_0^{\mathrm{\infty }}{\left(\frac{1-{\mathrm{e}}^{-2\mathrm{x}}}{1+{\mathrm{e}}^{-2\mathrm{x}}}\right)}^{\mathrm{a}}\mathrm{dx}{=}_{{\mathrm{e}}^{2\mathrm{x}}=\mathrm{t}}{\int }_1^{+\mathrm{\infty }}{\left(\frac{1-{\mathrm{t}}^{-1}}{1+{\mathrm{t}}^{-1}}\right)}^{\mathrm{a}}\frac{\mathrm{dt}}{2\mathrm{t}}$$$$=\frac{1}{2}{\int }_1^{\mathrm{\infty }}{\left(\frac{\mathrm{t}-1}{\mathrm{t}+1}\right)}^{\mathrm{a}}\frac{\mathrm{dt}}{\mathrm{t}}\mathrm{changement}\frac{\mathrm{t}-1}{\mathrm{t}+1}=\mathrm{u}\mathrm{givet}-1=\mathrm{ut}+\mathrm{u}\Rightarrow$$$$(1-\mathrm{u})\mathrm{t}=1+\mathrm{u}\Rightarrow \mathrm{t}=\frac{1+\mathrm{u}}{1-\mathrm{u}}\Rightarrow \frac{\mathrm{dt}}{\mathrm{du}}=\frac{1-\mathrm{u}-(1+\mathrm{u})(-1)}{{(1-\mathrm{u})}^2}=\frac{2}{{(1-\mathrm{u})}^2}$$$${\int }_1^{+\mathrm{\infty }}\frac{1}{\mathrm{t}}{\left(\frac{\mathrm{t}-1}{\mathrm{t}+1}\right)}^{\mathrm{a}}\mathrm{dt}={\int }_0^1{\mathrm{u}}^{\mathrm{a}}\times \frac{1-\mathrm{u}}{1+\mathrm{u}}\times \frac{2\mathrm{du}}{{(1-\mathrm{u})}^2}$$$$=2{\int }_0^1\frac{{\mathrm{u}}^{\mathrm{a}}}{1-{\mathrm{u}}^2}\mathrm{du}=2{\int }_0^1{\mathrm{u}}^{\mathrm{a}}\left(\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\mathrm{u}}^{2\mathrm{n}}\right)\mathrm{du}$$$$=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}{\int }_0^1{\mathrm{u}}^{2\mathrm{n}+\mathrm{a}}\mathrm{du}=2\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2\mathrm{n}+\mathrm{a}+1}\Rightarrow$$$$\phi \left(\mathrm{a}\right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2\mathrm{n}+1+\mathrm{a}}\Rightarrow \mathrm{I}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2\mathrm{n}+1+\mathrm{a}}-\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{2\mathrm{n}+1+\mathrm{b}}$$$$=\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+\frac{\mathrm{a}+1}{2}}-\frac{1}{2}\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+\frac{\mathrm{b}+1}{2}}$$$$\mathrm{so}\mathrm{if}\mathrm{we}\mathrm{take}\psi \left(\lambda \right)=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{1}{\mathrm{n}+\lambda }\mathrm{we}\mathrm{get}\mathrm{I}=\frac{1}{2}(\psi \left(\frac{\mathrm{a}+1}{2}\right)-\psi \left(\frac{\mathrm{b}+1}{2}\right))$$

Commented by mnjuly1970 last updated on 29/Sep/20

$$thankyousir$$

Commented by Bird last updated on 29/Sep/20

$$youarewelcomesir$$

Answered by mnjuly1970 last updated on 29/Sep/20

Answered by mnjuly1970 last updated on 29/Sep/20

Answered by mnjuly1970 last updated on 29/Sep/20

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