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Question Number 2434 by Yozzi last updated on 20/Nov/15

Prove that for m=0,1,2,3,... lim_(x→−m) Γ(x)=∞.

$${Prove}\:{that}\:{for}\:{m}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{x}\rightarrow−{m}} {\mathrm{lim}}\Gamma\left({x}\right)=\infty. \\ $$

Answered by 123456 last updated on 25/Nov/15

Γ(x+1)=xΓ(x) so Γ(x)=((Γ(x+1))/x) Γ(x−1)=((Γ(x))/(x−1)) with this we can undestand why gamma function have simply poles at x∈Z_− ∪{0} however if you plot the gamma function lim_(x→0) Γ(x)=∄ because lim_(x→0^+ ) Γ(x)=+∞ lim_(x→0^− ) Γ(x)=−∞ similiar things hold for the remaining

$$\Gamma\left({x}+\mathrm{1}\right)={x}\Gamma\left({x}\right) \\ $$$$\mathrm{so} \\ $$$$\Gamma\left({x}\right)=\frac{\Gamma\left({x}+\mathrm{1}\right)}{{x}} \\ $$$$\Gamma\left({x}−\mathrm{1}\right)=\frac{\Gamma\left({x}\right)}{{x}−\mathrm{1}} \\ $$$$\mathrm{with}\:\mathrm{this}\:\mathrm{we}\:\mathrm{can}\:\mathrm{undestand}\:\mathrm{why}\: \\ $$$$\mathrm{gamma}\:\mathrm{function}\:\mathrm{have}\:\mathrm{simply}\:\mathrm{poles} \\ $$$$\mathrm{at}\:{x}\in\mathbb{Z}_{−} \cup\left\{\mathrm{0}\right\} \\ $$$$\mathrm{however}\:\mathrm{if}\:\mathrm{you}\:\mathrm{plot}\:\mathrm{the}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\Gamma\left({x}\right)=\nexists \\ $$$$\mathrm{because} \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\Gamma\left({x}\right)=+\infty \\ $$$$\underset{{x}\rightarrow\mathrm{0}^{−} } {\mathrm{lim}}\:\Gamma\left({x}\right)=−\infty \\ $$$$\mathrm{similiar}\:\mathrm{things}\:\mathrm{hold}\:\mathrm{for}\:\mathrm{the}\:\mathrm{remaining} \\ $$

Commented by prakash jain last updated on 25/Nov/15

Γ(0)

$$\Gamma\left(\mathrm{0}\right)\: \\ $$

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