Question Number 137275 by mnjuly1970 last updated on 31/Mar/21
![......complex analysis..... if , f(α,n,x)=(d^( n) /dx^n )(α^x ) , x∈C α∈C−{0} , n∈C−Z^− ∪{0} and g(n,x)=∫_0 ^( 1) f(α,n,x)dα then find the value of ... Ω=((Im(g(i,0)))/(Re(Γ(i)))) solution: g(n,x)=∫_0 ^( 1) (d^( n) /dx^(n ) )(α^x )dα =∫_0 ^( 1) (d^( n) /dx^n )(e^(xln(α)) )dα .....⟨∗⟩ (d^( n) /dx^n )(e^(xln(α)) )=(ln(α))^n α^x ⟨∗⟩→ ...g(n,x)=∫_0 ^( 1) (ln(α))^n α^x dα =_(α=e^(−y) ) ^(ln(α)=−y) ∫_0 ^( ∞) (−1)^n y^( n) e^(−yx) e^(−y) dy =(−1)^n ∫_0 ^( 1) y^n .e^(−y(1+x)) dy =^(y(1+x)=t) (−1)^n ∫_0 ^( 1) ((t^n e^(−t) )/((1+x)^(n+1) ))dt =e^(inπ) .(1/((1+x)^(n+1) )) .Γ(n+1) g(i,0)=e^(−π) .i.Γ(i) ......⟨∗∗⟩ Γ(i)∈C ⇒ Γ(i)=Re(Γ(i))+Im(Γ(i)) ⟨∗∗⟩→ ... g(i,0)=e^(−π) .i.[Re(Γ(i))+Im(Γ(i))] ∴ Ω=((e^(−π) Re(Γ(i)))/(Re(Γ(i)))) =e^(−π) ...✓✓](Q137275.png)
$$\:\:\:\:\:\:\:\:\:\:\:\:\:......{complex}\:\:{analysis}..... \\ $$$$\:\:\:\:{if}\:,\:\:{f}\left(\alpha,{n},{x}\right)=\frac{{d}^{\:{n}} }{{dx}^{{n}} }\left(\alpha^{{x}} \right)\:\:,\:{x}\in\mathbb{C} \\ $$$$\:\:\:\:\:\alpha\in\mathbb{C}−\left\{\mathrm{0}\right\}\:,\:{n}\in\mathbb{C}−\mathbb{Z}^{−} \cup\left\{\mathrm{0}\right\} \\ $$$$\:\:\:\:\:{and}\:\:{g}\left({n},{x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left(\alpha,{n},{x}\right){d}\alpha \\ $$$$\:\:\:\:\:{then}\:\:{find}\:\:{the}\:{value}\:{of}\:... \\ $$$$\:\:\:\:\:\:\Omega=\frac{{Im}\left({g}\left({i},\mathrm{0}\right)\right)}{{Re}\left(\Gamma\left({i}\right)\right)} \\ $$$$\:\:\:\:\:\:{solution}: \\ $$$$\:\:\:\:\:\:{g}\left({n},{x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{d}^{\:{n}} }{{dx}^{{n}\:} }\left(\alpha^{{x}} \right){d}\alpha \\ $$$$\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{d}^{\:{n}} }{{dx}^{{n}} }\left({e}^{{xln}\left(\alpha\right)} \right){d}\alpha\:.....\langle\ast\rangle \\ $$$$\:\:\:\:\:\:\:\frac{{d}^{\:{n}} }{{dx}^{{n}} }\left({e}^{{xln}\left(\alpha\right)} \right)=\left({ln}\left(\alpha\right)\right)^{{n}} \alpha^{{x}} \\ $$$$\:\:\:\langle\ast\rangle\rightarrow\:...{g}\left({n},{x}\right)=\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({ln}\left(\alpha\right)\right)^{{n}} \alpha^{{x}} {d}\alpha \\ $$$$\:\:\underset{\alpha={e}^{−{y}} } {\overset{{ln}\left(\alpha\right)=−{y}} {=}}\int_{\mathrm{0}} ^{\:\infty} \left(−\mathrm{1}\right)^{{n}} {y}^{\:{n}} {e}^{−{yx}} {e}^{−{y}} {dy} \\ $$$$\:\:\:\:\:=\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\:\mathrm{1}} {y}^{{n}} .{e}^{−{y}\left(\mathrm{1}+{x}\right)} {dy} \\ $$$$\:\:\:\:\overset{{y}\left(\mathrm{1}+{x}\right)={t}} {=}\left(−\mathrm{1}\right)^{{n}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{t}^{{n}} \:{e}^{−{t}} }{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} }{dt} \\ $$$$\:\:\:\:\:\:\:={e}^{{in}\pi} .\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)^{{n}+\mathrm{1}} }\:.\Gamma\left({n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:{g}\left({i},\mathrm{0}\right)={e}^{−\pi} .{i}.\Gamma\left({i}\right)\:......\langle\ast\ast\rangle \\ $$$$\:\:\:\Gamma\left({i}\right)\in\mathbb{C}\:\Rightarrow\:\:\Gamma\left({i}\right)={Re}\left(\Gamma\left({i}\right)\right)+{Im}\left(\Gamma\left({i}\right)\right) \\ $$$$\:\:\:\langle\ast\ast\rangle\rightarrow\:...\:{g}\left({i},\mathrm{0}\right)={e}^{−\pi} .{i}.\left[{Re}\left(\Gamma\left({i}\right)\right)+{Im}\left(\Gamma\left({i}\right)\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\Omega=\frac{{e}^{−\pi} {Re}\left(\Gamma\left({i}\right)\right)}{{Re}\left(\Gamma\left({i}\right)\right)}\:={e}^{−\pi} ...\checkmark\checkmark \\ $$$$\:\:\:\:\: \\ $$$$\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$