......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) ...✓✓


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Question Number 137275 by mnjuly1970 last updated on 31/Mar/21
$......complex analysis..... if , f(α,n,x)=(d^( n) /dx^n )(α^x ) , x∈C α∈C−{0} , n∈C−Z^− ∪{0} and g(n,x)=∫_0 ^( 1) f(α,n,x)dα then find the value of ... Ω=((Im(g(i,0)))/(Re(Γ(i)))) solution: g(n,x)=∫_0 ^( 1) (d^( n) /dx^(n ) )(α^x )dα =∫_0 ^( 1) (d^( n) /dx^n )(e^(xln(α)) )dα .....⟨∗⟩ (d^( n) /dx^n )(e^(xln(α)) )=(ln(α))^n α^x ⟨∗⟩→ ...g(n,x)=∫_0 ^( 1) (ln(α))^n α^x dα =_(α=e^(−y) ) ^(ln(α)=−y) ∫_0 ^( ∞) (−1)^n y^( n) e^(−yx) e^(−y) dy =(−1)^n ∫_0 ^( 1) y^n .e^(−y(1+x)) dy =^(y(1+x)=t) (−1)^n ∫_0 ^( 1) ((t^n e^(−t) )/((1+x)^(n+1) ))dt =e^(inπ) .(1/((1+x)^(n+1) )) .Γ(n+1) g(i,0)=e^(−π) .i.Γ(i) ......⟨∗∗⟩ Γ(i)∈C ⇒ Γ(i)=Re(Γ(i))+Im(Γ(i)) ⟨∗∗⟩→ ... g(i,0)=e^(−π) .i.[Re(Γ(i))+Im(Γ(i))] ∴ Ω=((e^(−π) Re(Γ(i)))/(Re(Γ(i)))) =e^(−π) ...✓✓$
$. . . . . . c o m p l e x a n a l y s i s . . . . .$ $i f, f (α, n, x) = \frac{d^{n}}{{d x}^{n}} (α^{x}), x \in C$ $α \in C - {0}, n \in C - Z^{-} \cup {0}$ $a n d g (n, x) = \int_{0}^{1} f (α, n, x) d α$ $t h e n f i n d t h e v a l u e o f . . .$ $Ω = \frac{I m (g (i, 0))}{R e (Γ (i))}$ $s o l u t i o n :$ $g (n, x) = \int_{0}^{1} \frac{d^{n}}{{d x}^{n}} (α^{x}) d α$ $= \int_{0}^{1} \frac{d^{n}}{{d x}^{n}} (e^{x l n (α)}) d α . . . . . ⟨ * ⟩$ $\frac{d^{n}}{{d x}^{n}} (e^{x l n (α)}) = {(l n (α))}^{n} α^{x}$ $⟨ * ⟩ \to . . . g (n, x) = \int_{0}^{1} {(l n (α))}^{n} α^{x} d α$ $\underset{α = e^{- y}}{\overset{l n (α) = - y}{=}} \int_{0}^{\infty} {(- 1)}^{n} y^{n} e^{- y x} e^{- y} d y$ $= {(- 1)}^{n} \int_{0}^{1} y^{n} . e^{- y (1 + x)} d y$ $\overset{y (1 + x) = t}{=} {(- 1)}^{n} \int_{0}^{1} \frac{t^{n} e^{- t}}{{(1 + x)}^{n + 1}} d t$ $= e^{i n π} . \frac{1}{{(1 + x)}^{n + 1}} . Γ (n + 1)$ $g (i, 0) = e^{- π} . i . Γ (i) . . . . . . ⟨ * * ⟩$ $Γ (i) \in C \Rightarrow Γ (i) = R e (Γ (i)) + I m (Γ (i))$ $⟨ * * ⟩ \to . . . g (i, 0) = e^{- π} . i . [R e (Γ (i)) + I m (Γ (i))]$ $∴ Ω = \frac{e^{- π} R e (Γ (i))}{R e (Γ (i))} = e^{- π} . . . ✓ ✓$
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