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Question Number 115193 by mnjuly1970 last updated on 24/Sep/20

$$...advancedmathematics...$$$$::digammalimit::$$$$ifk>0then$$$$provethat$$$$$$$${lim}_{x\to 0}\frac{1}{x}(\psi \left(\frac{k+x}{2x}\right)-\psi \left(\frac{k}{2x}\right))=\frac{1}{k}✓$$$$$$$$m.n.july.1970...$$$$$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}$$

Answered by mathdave last updated on 24/Sep/20

$$solution$$$$letI=\underset{x\to 0}{\lim }(\frac{1}{x}(\psi (\frac{m}{2x}+\frac{1}{2})-\psi \left(\frac{m}{2x}\right))$$$$weknown$$$${\int }_0^1\frac{t^{n-1}}{1+t}dt=\frac{1}{2}(\psi (\frac{k}{2}+\frac{1}{2})-\psi \left(\frac{k}{2}\right))$$$$letk=\frac{m}{x}$$$$I=\frac{2}{x}{\int }_0^1\frac{t^{\frac{m}{x}-1}}{1+t}dtz=\frac{m}{x}$$$$I=\frac{2}{m}{\int }_0^1\frac{{zt}^{z-1}}{1+t}dt(let\int dv=\int {zt}^{z-1}dz,v=t^z$$$$andu=\frac{1}{1+t},du=-\frac{1}{{(1+t)}^2})usingIBP$$$$I=\frac{2}{m}{\left(\frac{t^z}{1+t}\right)}_0^1+\frac{2}{m}{\int }_0^1\frac{t^z}{{(1+t)}^2}dt=\frac{1}{m}+\frac{2}{m}\underset{z\to \mathrm{\infty }}{\lim }{\int }_0^1\frac{t^z}{{(1+t)}^2}dt$$$$lety=\frac{1}{t},dy=-\frac{1}{t^2}$$$$I=\frac{1}{m}+\frac{2}{m}\underset{z\to \mathrm{\infty }}{\lim }{\int }_{\mathrm{\infty }}^1\frac{y^{-z}}{{(1+\frac{1}{y})}^2}\times -\frac{1}{y^2}dy$$$$I=\frac{1}{m}+\frac{2}{m}\underset{z\to \mathrm{\infty }}{\lim }{\int }_0^{\mathrm{\infty }}\frac{y^2}{y^z{(1+y)}^2}\times \frac{dy}{y^2}=(\frac{1}{m}+\frac{2}{m}\underset{z\to \mathrm{\infty }}{\lim }{\int }_1^{\mathrm{\infty }}\frac{dy}{y^z{(1+y)}^2})=\frac{1}{m}$$$$\because \underset{x\to 0}{\lim }\frac{1}{x}(\psi (\frac{m}{2x}+\frac{1}{2})-\psi \left(\frac{m}{2x}\right))=\frac{1}{m}Q.E.D$$$$bymathdave\left(24/09/2020\right)$$$$$$

Commented by mnjuly1970 last updated on 24/Sep/20

$$goodwork$$

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