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Category: Relation and Functions

given-z-lim-n-n-z-z-1-z-n-n-z-proof-that-i-z-1-z-z-ii-wiertrass-definition-of-gamma-function-1-z-ze-z-m-1-1-z-m-e-z-m-is-euler-macheroni-con

Question Number 2891 by 123456 last updated on 29/Nov/15 $$\mathrm{given} \\ $$$$\Gamma\left({z}\right)=\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\:\frac{{n}!}{{z}\left({z}+\mathrm{1}\right)\centerdot\centerdot\centerdot\left({z}+{n}\right)}{n}^{{z}} \\ $$$$\mathrm{proof}\:\mathrm{that} \\ $$$$\left.{i}\right) \\ $$$$\Gamma\left({z}+\mathrm{1}\right)={z}\Gamma\left({z}\right) \\ $$$$\left.{ii}\right)\:\mathrm{wiertrass}\:\mathrm{definition}\:\mathrm{of}\:\mathrm{gamma}\:\mathrm{function} \\ $$$$\frac{\mathrm{1}}{\Gamma\left({z}\right)}={ze}^{{z}\gamma} \underset{{m}=\mathrm{1}} {\overset{+\infty}…

Analytical-Continuation-Sum-of-the-below-divergent-series-was-shown-to-be-using-analytical-continuation-i-1-n-2-i-1-1-A-1-i-0-i-1-12-B-While-reading-about

Question Number 2720 by prakash jain last updated on 25/Nov/15 $$\mathrm{Analytical}\:\mathrm{Continuation} \\ $$$$\mathrm{Sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{below}\:\mathrm{divergent}\:\mathrm{series}\:\mathrm{was} \\ $$$$\mathrm{shown}\:\mathrm{to}\:\mathrm{be}\:\mathrm{using}\:\mathrm{analytical}\:\mathrm{continuation}. \\ $$$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{2}^{{i}−\mathrm{1}} =−\mathrm{1}\:\:\:\:\:\:…\left(\mathrm{A}\right) \\ $$$$\zeta\left(−\mathrm{1}\right)=\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}{i}=−\frac{\mathrm{1}}{\mathrm{12}}\:\:\:\:\:\:…\left(\mathrm{B}\right) \\…

let-f-x-arctan-ax-1-with-a-real-1-calculate-f-n-x-and-f-n-0-2-developp-f-at-integr-serie-3-calculate-f-x-x-2-4-dx-

Question Number 68243 by mathmax by abdo last updated on 07/Sep/19 $${let}\:{f}\left({x}\right)\:={arctan}\left({ax}\:+\mathrm{1}\right)\:\:{with}\:{a}\:{real} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right)\:{and}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:{developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{f}\left({x}\right)}{{x}^{\mathrm{2}} \:+\mathrm{4}}{dx} \\ $$…