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Question Number 127315 by mnjuly1970 last updated on 28/Dec/20

...advanced calculud... compute ::: ψ(i)=??

$$\:\:\:\:\:\:\:\:\:...{advanced}\:\:\:{calculud}... \\ $$$$\:\:\:\:{compute}\:\:::: \\ $$$$\:\:\:\:\:\psi\left({i}\right)=?? \\ $$$$\:\:\:\:\:\:\: \\ $$

Commented by Dwaipayan Shikari last updated on 28/Dec/20

ψ(i)=−γ+∫_0 ^1 ((1−x^(i−1) )/(1−x))dx =−γ+∫_0 ^1 ((x−x^i )/x)+((x−x^i )/(1−x))dx = −γ+1+i+∫_0 ^1 ((x−x^i )/(1−x))dx I(a)=∫_0 ^1 ((x−x^i )/(1−x))dx=∫_0 ^1 (((1−x−(1−x)^i )/x))e^(−ax) dx I′(a)=∫_0 ^1 (1−x)^i e^(−ax) dx=∫_0 ^1 x^i e^(−a(1−x)) dx=e^(−a) ∫_0 ^1 x^i e^x dx =e^(−a) Σ_(n=0) ^∞ (1/(n!(n+i))) I(a)=−e^(−a) Σ_(n=1) ^∞ (1/(n!(n+i)))+C⇒I(0)=−Σ_(n=0) ^∞ (1/(n!(n+i))) ψ(i)=1−γ+i−Σ_(n=0) ^∞ (1/(n!(n+i)))....

$$\psi\left({i}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{x}^{{i}−\mathrm{1}} }{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{x}^{{i}} }{{x}}+\frac{{x}−{x}^{{i}} }{\mathrm{1}−{x}}{dx} \\ $$$$\:\:\:\:\:\:\:\:=\:−\gamma+\mathrm{1}+{i}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{x}^{{i}} }{\mathrm{1}−{x}}{dx} \\ $$$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−{x}^{{i}} }{\mathrm{1}−{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}−{x}−\left(\mathrm{1}−{x}\right)^{{i}} }{{x}}\right){e}^{−{ax}} {dx} \\ $$$${I}'\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{x}\right)^{{i}} {e}^{−{ax}} {dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{i}} {e}^{−{a}\left(\mathrm{1}−{x}\right)} {dx}={e}^{−{a}} \int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{i}} {e}^{{x}} {dx} \\ $$$$={e}^{−{a}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left({n}+{i}\right)} \\ $$$${I}\left({a}\right)=−{e}^{−{a}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left({n}+{i}\right)}+{C}\Rightarrow{I}\left(\mathrm{0}\right)=−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left({n}+{i}\right)} \\ $$$$\psi\left({i}\right)=\mathrm{1}−\gamma+{i}−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}!\left({n}+{i}\right)}.... \\ $$

Commented by dsardor2004 last updated on 28/Dec/20

?

$$? \\ $$

Answered by Olaf last updated on 28/Dec/20

Using series : Re(ψ(i)) = −γ−Σ_(n=0) ^∞ ((n−1)/(n^3 +n^2 +n+1)) Re(ψ(i)) = −γ+1+Σ_(n=2) ^∞ (1/(n^4 −1)) Im(ψ(i)) = Σ_(n=0) ^∞ (1/(n^2 +1)) = (1/2)+(π/2)cotπ ψ(i) = −γ+1+Σ_(n=2) ^∞ (1/(n^4 −1))+i((1+πcotπ)/2)

$${Using}\:{series}\:: \\ $$$$\mathrm{Re}\left(\psi\left({i}\right)\right)\:=\:−\gamma−\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{n}−\mathrm{1}}{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$$$\mathrm{Re}\left(\psi\left({i}\right)\right)\:=\:−\gamma+\mathrm{1}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} −\mathrm{1}} \\ $$$$\mathrm{Im}\left(\psi\left({i}\right)\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\mathrm{cot}\pi \\ $$$$\psi\left({i}\right)\:=\:−\gamma+\mathrm{1}+\underset{{n}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{4}} −\mathrm{1}}+{i}\frac{\mathrm{1}+\pi\mathrm{cot}\pi}{\mathrm{2}} \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 28/Dec/20

thank you ...

$${thank}\:{you}\:... \\ $$

Answered by mindispower last updated on 28/Dec/20

Ψ(i)=−γ+Σ_(n≥1) ((1/(n+1))−(1/(n+i))) =−γ+Σ_(n≥0) ((1/(n+1))−((n−i)/(n^2 +1))) =−γ+Σ_(n≥0) ((−n+1)/(n^3 +n^2 +n+1))+iΣ_(n≥0) (1/(n^2 +1)) ReΨ(i)=−γ−Σ_(n≥0) ((n−1)/(n^3 +n^2 +n+1)) Im(Ψ(i))=Σ_(n≥0) (1/(n^2 +1))=(1/2)(Σ_(−∞) ^∞ (1/(n^2 +1))+1) =(1/2)(1−2Reπcoth(iπ).(1/(1−i))) =(1/2)(1−Re(πicoth(π).(1+i)) =(1/2)(1+πcotth(π)) Im(Ψ(i))=(1/2)+(π/2)coth(π) Re(Ψ(i))=−γ−Σ_(n≥1) ((n−1)/(n^3 +n^2 +n+1))..no close form in elementry function

$$\Psi\left({i}\right)=−\gamma+\underset{{n}\geqslant\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{\mathrm{1}}{{n}+{i}}\right) \\ $$$$=−\gamma+\underset{{n}\geqslant\mathrm{0}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}−\frac{{n}−{i}}{{n}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$=−\gamma+\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{−{n}+\mathrm{1}}{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}+{i}\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}} \\ $$$${Re}\Psi\left({i}\right)=−\gamma−\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{{n}−\mathrm{1}}{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}} \\ $$$${Im}\left(\Psi\left({i}\right)\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\underset{−\infty} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}+\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{2}{Re}\pi{coth}\left({i}\pi\right).\frac{\mathrm{1}}{\mathrm{1}−{i}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−{Re}\left(\pi{icoth}\left(\pi\right).\left(\mathrm{1}+{i}\right)\right)\right. \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\pi{cotth}\left(\pi\right)\right) \\ $$$${Im}\left(\Psi\left({i}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}{coth}\left(\pi\right) \\ $$$${Re}\left(\Psi\left({i}\right)\right)=−\gamma−\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{{n}−\mathrm{1}}{{n}^{\mathrm{3}} +{n}^{\mathrm{2}} +{n}+\mathrm{1}}..{no}\:{close}\:{form}\: \\ $$$${in}\:{elementry}\:{function} \\ $$

Commented by mnjuly1970 last updated on 28/Dec/20

thank you so much sir power..

$${thank}\:{you}\:{so}\:{much}\:{sir}\:{power}.. \\ $$

Commented by mindispower last updated on 29/Dec/20

always pleasur sir

$${always}\:{pleasur}\:{sir} \\ $$

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