Question Number 220976 by SdC355 last updated on 21/May/25 $$\int_{−\infty} ^{\:+\infty} \int_{−\infty} ^{\:+\infty} \:{e}^{−\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)} \mathrm{d}{V}\:\rightarrow\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \int_{\mathrm{0}} ^{\:\infty} \:{r}\centerdot{e}^{−{r}^{\mathrm{2}} } \mathrm{d}{r}\mathrm{d}\theta=\pi \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{understand}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{integration}…
Question Number 220972 by SdC355 last updated on 21/May/25 $$\int_{−\infty} ^{+\infty} \int_{−\infty} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{4}\right)^{\mathrm{2}} }\:\mathrm{d}{x}\mathrm{d}{y} \\ $$$${x}={r}\mathrm{cos}\left(\theta\right) \\ $$$${y}={r}\mathrm{sin}\left(\theta\right) \\ $$$$\mid\mid\boldsymbol{{J}}\mid\mid={r}\mathrm{d}{r}\mathrm{d}\theta \\ $$$$\int_{\mathrm{0}}…
Question Number 220973 by MrGaster last updated on 21/May/25 $$\mathrm{Prove}:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{k}} \mathrm{cos}^{\mathrm{3}} \left(\mathrm{3}^{{k}−{n}} \pi\right)=\frac{\mathrm{3}}{\mathrm{4}}\left[\left(−\frac{\mathrm{1}}{\mathrm{3}}\right)^{{n}+\mathrm{1}} +\mathrm{cos}\frac{\pi}{\mathrm{3}^{{n}} }\right] \\ $$ Answered by universe last updated on…
Question Number 220913 by SdC355 last updated on 20/May/25 $$\mathrm{Is}\:\mathrm{there}\:\mathrm{an}\:\mathrm{Manager}??? \\ $$$$\mathrm{pls}\:\mathrm{ban}\:\mathrm{Question}\:\mathrm{Spamming}\:\mathrm{and}… \\ $$$$\mathrm{pls}\:\mathrm{fix}\:\mathrm{invisible}\:\mathrm{line}\:\mathrm{matrix}\:\mathrm{bug} \\ $$ Commented by mr W last updated on 21/May/25 $${the}\:{developer}\:{Tinku}\:{Tara}\:{doesn}'{t}…
Question Number 220820 by SdC355 last updated on 19/May/25 Commented by SdC355 last updated on 19/May/25 $$\mathrm{Q220800} \\ $$ Answered by MathematicalUser2357 last updated on…
Question Number 220790 by SdC355 last updated on 19/May/25 $$\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…
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…
Question Number 220755 by SdC355 last updated on 18/May/25 $$ \\ $$$$\mathrm{Hmm}….\mathrm{i}\:\mathrm{need}\:\mathrm{a}\:\mathrm{help}… \\ $$$$\mathrm{generalize}\: \\ $$$$\mathcal{L}_{{z}} ^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{{z}^{{n}} +\mathrm{1}}\right\}=\:\oint_{\:{C}} \:\:\frac{{e}^{{zt}} }{{z}^{{n}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$ Terms…
Question Number 220664 by SdC355 last updated on 17/May/25 $$\:\mathrm{example}.. \\ $$$${x}\left({t}\right)=\oint_{\:{C}} \:\:\frac{{s}−\mathrm{2}}{{s}^{\mathrm{2}} −\mathrm{4}{s}+\mathrm{6}}\:{e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{Res}_{{s}=\mathrm{2}+\sqrt{\mathrm{2}}\boldsymbol{{i}}} \left\{\frac{{s}−\mathrm{2}}{{s}−\mathrm{2}+\sqrt{\mathrm{2}}\boldsymbol{{i}}}\right\}{e}^{{st}} +\mathrm{Res}_{{s}=\mathrm{2}−\sqrt{\mathrm{2}}\boldsymbol{{i}}} \left\{\frac{{s}−\mathrm{2}}{{s}−\mathrm{2}−\sqrt{\mathrm{2}}\boldsymbol{{i}}}\right\}{e}^{{st}} \\ $$$$\mathrm{because}\:\Sigma\:\mathrm{Res}_{{s}={z}_{{j}} } \left\{{f}\left({s}\right)\right\}{e}^{{st}} \\…
Question Number 220660 by SdC355 last updated on 17/May/25 $$\mathrm{Hmmm}…… \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{explain}?? \\ $$$$\mathrm{Let}\:\boldsymbol{\omega}={u}\:\mathrm{d}{x}+{v}\:\mathrm{d}{y}\:\mathrm{be}\:\mathrm{a}\:\mathrm{1}-\mathrm{form}\:\mathrm{defined}\:\mathrm{over}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{By}\:\mathrm{applying}\:\mathrm{the}\:\mathrm{above}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{each}\:\mathrm{terms} \\ $$$$\mathrm{consider}\:{x}^{\mathrm{1}} ={u}\:,\:{x}^{\mathrm{2}} ={v} \\ $$$$\mathrm{d}\boldsymbol{\omega}=\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{x}\right)+\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}}…