Question Number 115193 by mnjuly1970 last updated on 24/Sep/20
$$\:\:\:\:\:\:\:\:\:\:\:…{advanced}\:\:{mathematics}…\:\: \\ $$$$\:\:\:\:\:\:\:::\:\:\:{digamma}\:\:{limit}\:\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:{if}\:\:\:{k}>\mathrm{0}\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{{x}}\left(\psi\left(\frac{{k}+{x}}{\mathrm{2}{x}}\right)\:−\:\psi\left(\frac{{k}}{\mathrm{2}{x}}\right)\right)\:=\frac{\mathrm{1}}{{k}}\:\:\:\:\checkmark \\ $$$$ \\ $$$$\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}… \\ $$$$\: \\ $$
Commented by Tawa11 last updated on 06/Sep/21
$$\mathrm{great} \\ $$
Answered by mathdave last updated on 24/Sep/20
$${solution} \\ $$$${let}\:{I}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}}\left(\psi\left(\frac{{m}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{m}}{\mathrm{2}{x}}\right)\right)\right. \\ $$$${we}\:{known} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{{k}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{k}}{\mathrm{2}}\right)\right) \\ $$$${let}\:{k}=\frac{{m}}{{x}} \\ $$$${I}=\frac{\mathrm{2}}{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{m}}{{x}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:\:\:{z}=\frac{{m}}{{x}} \\ $$$${I}=\frac{\mathrm{2}}{{m}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{zt}^{{z}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\left(\:{let}\:\:\int{dv}=\int{zt}^{{z}−\mathrm{1}} {dz},{v}={t}^{{z}} \:\right. \\ $$$$\left.{and}\:{u}=\frac{\mathrm{1}}{\mathrm{1}+{t}},{du}=−\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right)\:{using}\:{IBP} \\ $$$${I}=\frac{\mathrm{2}}{{m}}\left(\frac{{t}^{{z}} }{\mathrm{1}+{t}}\right)_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{2}}{{m}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{z}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{z}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$${let}\:{y}=\frac{\mathrm{1}}{{t}},{dy}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$$${I}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\infty} ^{\mathrm{1}} \frac{{y}^{−{z}} }{\left(\mathrm{1}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} }×−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }{dy} \\ $$$${I}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\infty} \frac{{y}^{\mathrm{2}} }{{y}^{{z}} \left(\mathrm{1}+{y}\right)^{\mathrm{2}} }×\frac{{dy}}{{y}^{\mathrm{2}} }=\left(\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{1}} ^{\infty} \frac{{dy}}{{y}^{{z}} \left(\mathrm{1}+{y}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{{m}} \\ $$$$\because\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\left(\psi\left(\frac{{m}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{m}}{\mathrm{2}{x}}\right)\right)=\frac{\mathrm{1}}{{m}}\:\:\:{Q}.{E}.{D} \\ $$$${by}\:{mathdave}\left(\mathrm{24}/\mathrm{09}/\mathrm{2020}\right) \\ $$$$ \\ $$
Commented by mnjuly1970 last updated on 24/Sep/20
$${good}\:{work} \\ $$