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Question Number 145197 by Gbenga last updated on 03/Jul/21

d/dx of x!=?

$${d}/{dx}\:{of}\:{x}!=? \\ $$

Commented by Dwaipayan Shikari last updated on 03/Jul/21

x!=Γ(x+1) ((dx!)/dx)=Γ′(x+1)=Γ(x+1)ψ(x+1) As ((Γ′(x+1))/(Γ(x+1)))=ψ(x+1)

$${x}!=\Gamma\left({x}+\mathrm{1}\right) \\ $$$$\frac{{dx}!}{{dx}}=\Gamma'\left({x}+\mathrm{1}\right)=\Gamma\left({x}+\mathrm{1}\right)\psi\left({x}+\mathrm{1}\right)\:\:\:\:\:\:{As}\:\frac{\Gamma'\left({x}+\mathrm{1}\right)}{\Gamma\left({x}+\mathrm{1}\right)}=\psi\left({x}+\mathrm{1}\right) \\ $$

Answered by puissant last updated on 03/Jul/21

Γ(x)=(x−1)!=∫_0 ^(+∞) t^(x−1) e^(−t) dt (1/(Γ(x)))= xe^(γx) Π_(k=1) ^∞ (1+(x/k))e^(−(x/k)) ln((1/(Γ(x))))=−ln(Γ(x)) =ln(x)+γx+Σ_(k=1) ^∞ (1+(x/k))−(x/k) (d/dx)(ln((1/(Γ(x)))))=−(ln(Γ(x)))′ =− ((Γ′(x))/(Γ(x))) =(1/x)+γ+Σ_(k=1) ^∞ (1/(1+(x/k)))×(1/k)−(1/k) =−(1/x)−γ+Σ_(k=1) ^∞ (1/k)−(1/(x+k)) λ(x)=−γ+Σ_(k=1) ^∞ (1/k)−(1/(k−1+x)) ⇒λ(x+1)=−γ+Σ_(k=1) ^∞ (1/k)−(1/(k+x)) ⇒(d/dx)(x!)=((Γ′(x+1))/(Γ(x+1)))....

$$\Gamma\left(\mathrm{x}\right)=\left(\mathrm{x}−\mathrm{1}\right)!=\int_{\mathrm{0}} ^{+\infty} \mathrm{t}^{\mathrm{x}−\mathrm{1}} \mathrm{e}^{−\mathrm{t}} \mathrm{dt} \\ $$$$\frac{\mathrm{1}}{\Gamma\left(\mathrm{x}\right)}=\:\mathrm{xe}^{\gamma\mathrm{x}} \underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{k}}\right)\mathrm{e}^{−\frac{\mathrm{x}}{\mathrm{k}}} \\ $$$$\mathrm{ln}\left(\frac{\mathrm{1}}{\Gamma\left(\mathrm{x}\right)}\right)=−\mathrm{ln}\left(\Gamma\left(\mathrm{x}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{ln}\left(\mathrm{x}\right)+\gamma\mathrm{x}+\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\mathrm{1}+\frac{\mathrm{x}}{\mathrm{k}}\right)−\frac{\mathrm{x}}{\mathrm{k}} \\ $$$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ln}\left(\frac{\mathrm{1}}{\Gamma\left(\mathrm{x}\right)}\right)\right)=−\left(\mathrm{ln}\left(\Gamma\left(\mathrm{x}\right)\right)\right)'\:=−\:\frac{\Gamma'\left(\mathrm{x}\right)}{\Gamma\left(\mathrm{x}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{x}}+\gamma+\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{x}}{\mathrm{k}}}×\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{x}}−\gamma+\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{x}+\mathrm{k}} \\ $$$$\lambda\left(\mathrm{x}\right)=−\gamma+\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}−\mathrm{1}+\mathrm{x}} \\ $$$$\Rightarrow\lambda\left(\mathrm{x}+\mathrm{1}\right)=−\gamma+\underset{\mathrm{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{k}}−\frac{\mathrm{1}}{\mathrm{k}+\mathrm{x}} \\ $$$$\Rightarrow\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}!\right)=\frac{\Gamma'\left(\mathrm{x}+\mathrm{1}\right)}{\Gamma\left(\mathrm{x}+\mathrm{1}\right)}.... \\ $$

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