Question Number 206674 by 073 last updated on 22/Apr/24 Answered by Frix last updated on 22/Apr/24 $$\Gamma\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}} \mathrm{e}^{−{t}} {dt} \\ $$$$\frac{{d}\Gamma\left({x}\right)}{{dx}}=\Gamma'\left({x}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{{x}−\mathrm{1}}…
Question Number 206600 by naka3546 last updated on 20/Apr/24 Answered by aleks041103 last updated on 22/Apr/24 $$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}\overset{\mathrm{3}{rd}\:{row}\:+\:\mathrm{1}{st}\:{row}} {\rightarrow}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{1}}&{−\mathrm{3}}&{−\mathrm{2}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{−\mathrm{1}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\overset{\mathrm{1}{st}\:{row}\:+\:\mathrm{2}{nd}\:{row}} {\rightarrow}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix} \\ $$$$\Rightarrow\begin{bmatrix}{−\mathrm{1}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}=\begin{bmatrix}{\mathrm{1}}&{−\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{1}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}\begin{bmatrix}{\mathrm{0}}&{\mathrm{3}}&{\mathrm{3}}\\{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{bmatrix}…
Question Number 206629 by aras18 last updated on 20/Apr/24 Answered by Frix last updated on 20/Apr/24 $${a}+{b}−{c}=\mathrm{170} \\ $$$${a}−{b}+{c}=\mathrm{130} \\ $$$$======= \\ $$$$\mathrm{2}{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{300} \\ $$$$\:\:\:{a}\:\:\:\:\:\:\:\:\:\:\:=\mathrm{150}…
Question Number 206550 by SANOGO last updated on 18/Apr/24 $${calcul}\:\:{Residus}\:{de}\:{f}\:{en}\:{o} \\ $$$${f}\left({z}\right)=\frac{{ze}^{{z}} }{\left({z}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$ Answered by Berbere last updated on 18/Apr/24 $${f}\:\:{n}'{a}\:{pas}\:{de}\:{poles}\:{en}\:\mathrm{0};{Res}\left({f},\mathrm{0}\right)=\mathrm{0} \\…
Question Number 206511 by naka3546 last updated on 17/Apr/24 Commented by mr W last updated on 17/Apr/24 $${i}\:{got}\:\mathrm{17178}. \\ $$ Commented by A5T last updated…
Question Number 206522 by MrGHK last updated on 17/Apr/24 Answered by Berbere last updated on 17/Apr/24 $${u}'={xln}^{\mathrm{2}} \left({x}\right)\Rightarrow{u}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} {ln}^{\mathrm{2}} \left({x}\right)−{x}^{\mathrm{2}} {ln}\left({x}\right)+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$${v}={Li}_{\mathrm{2}} \left(\frac{\mathrm{1}+{x}}{{x}}\right);{v}'=−\frac{\mathrm{1}}{{x}^{\mathrm{2}}…
Question Number 206484 by sonukgindia last updated on 16/Apr/24 Answered by mr W last updated on 16/Apr/24 Commented by mr W last updated on 16/Apr/24…
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Question Number 206393 by MaruMaru last updated on 13/Apr/24 $$\mathrm{find}\:\mathrm{S}=\mathrm{1}+\underset{\ell} {\sum}\:\frac{\left(−\right)^{\ell} }{\ell}\left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}\right)\:,\:\ell\in\left[\mathrm{1},\infty\right) \\ $$$$\mathrm{1}+\underset{\ell} {\sum}\:\frac{\left(−\right)^{\ell} }{\ell}\left(\frac{\mathrm{1}}{\ell}−\frac{\mathrm{1}}{\ell+\mathrm{1}}\right) \\ $$$$\mathrm{1}−\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\right)−\frac{\mathrm{1}}{\mathrm{3}}\left(\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{5}}\right)−…… \\ $$ Answered by MaruMaru last updated…
Question Number 206364 by Skabetix last updated on 12/Apr/24 Answered by TonyCWX08 last updated on 13/Apr/24 $${I}\:{only}\:{know} \\ $$$${e}^{\pi{i}} =−\mathrm{1} \\ $$$$\underset{−\infty} {\overset{\infty} {\int}}\frac{\mathrm{sin}\:\left({x}\right)}{{x}}\:{dx}\:=\:\pi \\…