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Author: Tinku Tara

Complex-integral-C-dz-z-3-1-C-x-2-y-2-4-C-1-z-e-z-dz-C-y-1-1-x-1-y-1-1-x-1-x-1-1-y-1-x-1-1-y-1-

Question Number 220790 by SdC355 last updated on 19/May/25 $$\mathrm{Complex}\:\mathrm{integral} \\ $$$$\oint_{\:\mathrm{C}} \:\frac{\mathrm{d}{z}}{{z}^{\mathrm{3}} +\mathrm{1}}=??\:,\:{C};{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{4} \\ $$$$\oint_{\:{C}} \:\frac{\mathrm{1}}{{z}}{e}^{{z}} \:\mathrm{d}{z},\:{C};\begin{cases}{{y}=\mathrm{1}\:,\:−\mathrm{1}\leq{x}\leq\mathrm{1}}\\{{y}=−\mathrm{1}\:,\:−\mathrm{1}\leq{x}\leq\mathrm{1}}\\{{x}=\mathrm{1}\:,\:−\mathrm{1}\leq{y}\leq\mathrm{1}}\\{{x}=−\mathrm{1}\:,\:−\mathrm{1}\leq{y}\leq\mathrm{1}}\end{cases}\: \\ $$ Answered by vnm…

To-Tinkutara-a-b-determinant-a-b-a-b-a-b-a-b-determinant-abcdefg-pqrstvw-is-work-well-but-invisible-line-matrix-dosen-t-work-pls-Fix-bug-

Question Number 220800 by SdC355 last updated on 19/May/25 $$\mathrm{To}\:\mathrm{Tinkutara} \\ $$$$\begin{bmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{bmatrix},\begin{vmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{vmatrix},\begin{pmatrix}{\mathrm{a}}\\{\mathrm{b}}\end{pmatrix}\:,\begin{cases}{\mathrm{a}}\\{\mathrm{b}}\end{cases}\:,\:\:\left.\begin{matrix}{\mathrm{a}}\\{\mathrm{b}}\end{matrix}\right\}\:,\begin{array}{|c|c|}{\mathrm{abcdefg}}\\{\mathrm{pqrstvw}}\\\hline\end{array}\:\mathrm{is}\:\mathrm{work}\:\mathrm{well} \\ $$$$\mathrm{but}\:\mathrm{invisible}\:\mathrm{line}\:\mathrm{matrix}\left(?\right)\:\mathrm{dosen}'\mathrm{t}\:\mathrm{work} \\ $$$$\mathrm{pls}\:\mathrm{Fix}\:\mathrm{bug} \\ $$ Answered by SdC355 last updated on 19/May/25…

L-lim-n-k-1-n-k-n-2-k-2-1-0-e-x-2-dx-1-m-0-1-m-2m-1-3-m-

Question Number 220764 by Nicholas666 last updated on 18/May/25 $$ \\ $$$$\:\:\boldsymbol{\mathrm{L}}=\:\boldsymbol{\mathrm{lim}}\underset{\:\boldsymbol{{n}}\rightarrow\infty} {\:}\left(\underset{\boldsymbol{{k}}=\mathrm{1}} {\overset{\boldsymbol{{n}}} {\sum}}\:\frac{\boldsymbol{{k}}}{\boldsymbol{{n}}^{\mathrm{2}} +\boldsymbol{{k}}^{\mathrm{2}} }\right).\left(\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \boldsymbol{{e}}^{−\boldsymbol{{x}}^{\mathrm{2}} } \boldsymbol{{dx}}\overset{−\mathrm{1}} {\right)}.\left(\underset{\boldsymbol{{m}}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{{m}}} }{\left(\mathrm{2}\boldsymbol{{m}}+\mathrm{1}\right)\mathrm{3}^{\boldsymbol{{m}}}…