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

...advanced mathematics... :: digamma limit :: if k>0 then prove that lim_(x→0) (1/x)(ψ(((k+x)/(2x))) − ψ((k/(2x)))) =(1/k) ✓ m.n.july.1970...

$$\:\:\:\:\:\:\:\:\:\:\:...{advanced}\:\:{mathematics}...\:\: \\ $$ $$\:\:\:\:\:\:\:::\:\:\:{digamma}\:\:{limit}\:\::: \\ $$ $$\:\:\:\:\:\:\:\:\:\:{if}\:\:\:{k}>\mathrm{0}\:\:{then} \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{prove}\:\:{that}\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{\mathrm{1}}{{x}}\left(\psi\left(\frac{{k}+{x}}{\mathrm{2}{x}}\right)\:−\:\psi\left(\frac{{k}}{\mathrm{2}{x}}\right)\right)\:=\frac{\mathrm{1}}{{k}}\:\:\:\:\checkmark \\ $$ $$ \\ $$ $$\:\:\:\:\:{m}.{n}.{july}.\mathrm{1970}... \\ $$ $$\: \\ $$

Commented byTawa11 last updated on 06/Sep/21

great

$$\mathrm{great} \\ $$

Answered by mathdave last updated on 24/Sep/20

solution let I=lim_(x→0) ((1/x)(ψ((m/(2x))+(1/2))−ψ((m/(2x)))) we known ∫_0 ^1 (t^(n−1) /(1+t))dt=(1/2)(ψ((k/2)+(1/2))−ψ((k/2))) let k=(m/x) I=(2/x)∫_0 ^1 (t^((m/x)−1) /(1+t))dt z=(m/x) I=(2/m)∫_0 ^1 ((zt^(z−1) )/(1+t))dt ( let ∫dv=∫zt^(z−1) dz,v=t^z and u=(1/(1+t)),du=−(1/((1+t)^2 ))) using IBP I=(2/m)((t^z /(1+t)))_0 ^1 +(2/m)∫_0 ^1 (t^z /((1+t)^2 ))dt=(1/m)+(2/m)lim_(z→∞) ∫_0 ^1 (t^z /((1+t)^2 ))dt let y=(1/t),dy=−(1/t^2 ) I=(1/m)+(2/m)lim_(z→∞) ∫_∞ ^1 (y^(−z) /((1+(1/y))^2 ))×−(1/y^2 )dy I=(1/m)+(2/m)lim_(z→∞) ∫_0 ^∞ (y^2 /(y^z (1+y)^2 ))×(dy/y^2 )=((1/m)+(2/m)lim_(z→∞) ∫_1 ^∞ (dy/(y^z (1+y)^2 )))=(1/m) ∵lim_(x→0) (1/x)(ψ((m/(2x))+(1/2))−ψ((m/(2x))))=(1/m) Q.E.D by mathdave(24/09/2020)

$${solution} \\ $$ $${let}\:{I}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{1}}{{x}}\left(\psi\left(\frac{{m}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{m}}{\mathrm{2}{x}}\right)\right)\right. \\ $$ $${we}\:{known} \\ $$ $$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{n}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{{k}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{k}}{\mathrm{2}}\right)\right) \\ $$ $${let}\:{k}=\frac{{m}}{{x}} \\ $$ $${I}=\frac{\mathrm{2}}{{x}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{\frac{{m}}{{x}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\:\:\:{z}=\frac{{m}}{{x}} \\ $$ $${I}=\frac{\mathrm{2}}{{m}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{zt}^{{z}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:\left(\:{let}\:\:\int{dv}=\int{zt}^{{z}−\mathrm{1}} {dz},{v}={t}^{{z}} \:\right. \\ $$ $$\left.{and}\:{u}=\frac{\mathrm{1}}{\mathrm{1}+{t}},{du}=−\frac{\mathrm{1}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }\right)\:{using}\:{IBP} \\ $$ $${I}=\frac{\mathrm{2}}{{m}}\left(\frac{{t}^{{z}} }{\mathrm{1}+{t}}\right)_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{2}}{{m}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{z}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{z}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$ $${let}\:{y}=\frac{\mathrm{1}}{{t}},{dy}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} } \\ $$ $${I}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\infty} ^{\mathrm{1}} \frac{{y}^{−{z}} }{\left(\mathrm{1}+\frac{\mathrm{1}}{{y}}\right)^{\mathrm{2}} }×−\frac{\mathrm{1}}{{y}^{\mathrm{2}} }{dy} \\ $$ $${I}=\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\infty} \frac{{y}^{\mathrm{2}} }{{y}^{{z}} \left(\mathrm{1}+{y}\right)^{\mathrm{2}} }×\frac{{dy}}{{y}^{\mathrm{2}} }=\left(\frac{\mathrm{1}}{{m}}+\frac{\mathrm{2}}{{m}}\underset{{z}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{1}} ^{\infty} \frac{{dy}}{{y}^{{z}} \left(\mathrm{1}+{y}\right)^{\mathrm{2}} }\right)=\frac{\mathrm{1}}{{m}} \\ $$ $$\because\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{1}}{{x}}\left(\psi\left(\frac{{m}}{\mathrm{2}{x}}+\frac{\mathrm{1}}{\mathrm{2}}\right)−\psi\left(\frac{{m}}{\mathrm{2}{x}}\right)\right)=\frac{\mathrm{1}}{{m}}\:\:\:{Q}.{E}.{D} \\ $$ $${by}\:{mathdave}\left(\mathrm{24}/\mathrm{09}/\mathrm{2020}\right) \\ $$ $$ \\ $$

Commented bymnjuly1970 last updated on 24/Sep/20

good work

$${good}\:{work} \\ $$

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