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Question Number 220741    Answers: 1   Comments: 0

Question Number 220740    Answers: 1   Comments: 0

Question Number 220739    Answers: 1   Comments: 0

Question Number 220738    Answers: 1   Comments: 0

Question Number 220737    Answers: 2   Comments: 0

Question Number 220730    Answers: 2   Comments: 0

∫_0 ^( 1) (1/( (√(x(1 − x)(1 + kx))))) dx , (−1 < k < 1)

$$ \\ $$$$\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{1}}{\:\sqrt{{x}\left(\mathrm{1}\:−\:{x}\right)\left(\mathrm{1}\:+\:{kx}\right)}}\:{dx}\:,\:\left(−\mathrm{1}\:<\:{k}\:<\:\mathrm{1}\right)\:\:\: \\ $$$$ \\ $$

Question Number 220726    Answers: 0   Comments: 4

Question Number 220715    Answers: 2   Comments: 1

Question Number 220712    Answers: 0   Comments: 0

prove: ((2 tan 2A + tan A)/(4 tan 3A − tan 2A)) = sin^2 A

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{prove}: \\ $$$$\:\:\frac{\mathrm{2}\:{tan}\:\mathrm{2}{A}\:+\:{tan}\:{A}}{\mathrm{4}\:{tan}\:\mathrm{3}{A}\:−\:{tan}\:\mathrm{2}{A}}\:=\:{sin}^{\mathrm{2}} \:{A} \\ $$$$\: \\ $$

Question Number 220707    Answers: 2   Comments: 0

∫_(−∞) ^(+∞) ((x(tan^(−1) x)^3 )/((x^2 +1)^2 (1+e^(4tan^(−1) x) )))dx=?

$$\underset{−\infty} {\overset{+\infty} {\int}}\frac{{x}\left(\mathrm{tan}^{−\mathrm{1}} \:{x}\right)^{\mathrm{3}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+\mathrm{e}^{\mathrm{4tan}^{−\mathrm{1}} \:{x}} \right)}{dx}=? \\ $$

Question Number 220706    Answers: 0   Comments: 0

Question Number 220704    Answers: 0   Comments: 0

Question Number 220700    Answers: 1   Comments: 0

Question Number 220694    Answers: 1   Comments: 1

Question Number 220693    Answers: 3   Comments: 2

Question Number 220677    Answers: 1   Comments: 0

∫(√(x+(√(x^2 +1 ))))dx

$$\int\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}\:}}{dx} \\ $$

Question Number 220676    Answers: 1   Comments: 0

∫ ((xdx)/((1−cosx)^2 ))

$$\int\:\frac{{xdx}}{\left(\mathrm{1}−{cosx}\right)^{\mathrm{2}} } \\ $$

Question Number 220674    Answers: 1   Comments: 0

∫∫∫_( E ) (z^2 /( (√(x^2 + y^2 )))) dV with the boundaries of the integration region E defined by; • x^2 + y^2 + z^2 ≤ 4 • x^2 + y^2 ≥ 1 • z ≥ 0

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\int\int_{\:{E}\:} \:\:\frac{{z}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} }}\:\:{dV} \\ $$$$\:\:\:\:\:\mathrm{with}\:\mathrm{the}\:\mathrm{boundaries}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integration}\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{region}\:{E}\:\mathrm{defined}\:\mathrm{by};\: \\ $$$$\:\:\:\:\:\:\bullet\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} +\:{z}^{\mathrm{2}} \:\leqslant\:\mathrm{4} \\ $$$$\:\:\:\:\:\:\bullet\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:\geqslant\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\bullet\:{z}\:\geqslant\:\mathrm{0} \\ $$$$ \\ $$

Question Number 220673    Answers: 0   Comments: 0

evaluate 1.∫_0 ^( ∞) ((cos(mt))/(t^2 +1)) dt 2. ∫_0 ^( ∞) sin(z^2 ) dz and ∫_0 ^( ∞) cos(z^2 ) dz by using complex integral..

$$\mathrm{evaluate} \\ $$$$\mathrm{1}.\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{cos}\left({mt}\right)}{{t}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{t} \\ $$$$\mathrm{2}.\:\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}\left({z}^{\mathrm{2}} \right)\:\mathrm{d}{z}\:\mathrm{and}\:\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{cos}\left({z}^{\mathrm{2}} \right)\:\mathrm{d}{z} \\ $$$${by}\:{using}\:{complex}\:{integral}.. \\ $$

Question Number 220664    Answers: 0   Comments: 0

example.. x(t)=∮_( C) ((s−2)/(s^2 −4s+6)) e^(st) ds Res_(s=2+(√2)i) {((s−2)/(s−2+(√2)i))}e^(st) +Res_(s=2−(√2)i) {((s−2)/(s−2−(√2)i))}e^(st) because Σ Res_(s=z_j ) {f(s)}e^(st) x(t)=(((√2)i)/(2(√2)i))e^(2t+(√2)it) +(((√2)i)/(2(√2)i))e^(2t−(√2)it) x(t)=e^(2t) (((e^((√2)it) +e^(−(√2)it) )/2)) ∴x(t)=e^(2t) cos((√2)t) Bromwich integral is defined as L_s ^(−1) {f(s)}= (1/(2πi)) ∮_( C) f(s)e^(st) ds if complex function f(s) is entire Does L_s ^(−1) {f(s)} dosen′t Exist? for example.... ∫_0 ^( ∞) J_ν (r)e^(−rt) dr=(((t+(√(t^2 +1)))^(−ν) )/( (√(t^2 +1)))) and we all know J_ν (s)=∮_( C) (((t+(√(t^2 +1)))^(−ν) )/( (√(t^2 +1)))) e^(st) dt But....can′t calculate...Σ_(h=1) ^N Res_(t=z_h ) {(((t+(√(t^2 +1)))^(−ν) )/( (√(t^2 +1))))} e^(st) and J_ν (s) can′t express by e^(λ_h t) , h=1,2,3... and by my Searching J_ν (s) defined as Σ_(j=0) ^∞ Res_(t=−j−(1/2)) {((𝚪(s+(1/2)ν))/(𝚪(1−s+(1/2)ν)))((s/2))^(2s) }

$$\:\mathrm{example}.. \\ $$$${x}\left({t}\right)=\oint_{\:{C}} \:\:\frac{{s}−\mathrm{2}}{{s}^{\mathrm{2}} −\mathrm{4}{s}+\mathrm{6}}\:{e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{Res}_{{s}=\mathrm{2}+\sqrt{\mathrm{2}}\boldsymbol{{i}}} \left\{\frac{{s}−\mathrm{2}}{{s}−\mathrm{2}+\sqrt{\mathrm{2}}\boldsymbol{{i}}}\right\}{e}^{{st}} +\mathrm{Res}_{{s}=\mathrm{2}−\sqrt{\mathrm{2}}\boldsymbol{{i}}} \left\{\frac{{s}−\mathrm{2}}{{s}−\mathrm{2}−\sqrt{\mathrm{2}}\boldsymbol{{i}}}\right\}{e}^{{st}} \\ $$$$\mathrm{because}\:\Sigma\:\mathrm{Res}_{{s}={z}_{{j}} } \left\{{f}\left({s}\right)\right\}{e}^{{st}} \\ $$$${x}\left({t}\right)=\frac{\sqrt{\mathrm{2}}\boldsymbol{{i}}}{\mathrm{2}\sqrt{\mathrm{2}}\boldsymbol{{i}}}{e}^{\mathrm{2}{t}+\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} +\frac{\sqrt{\mathrm{2}}\boldsymbol{{i}}}{\mathrm{2}\sqrt{\mathrm{2}}\boldsymbol{{i}}}{e}^{\mathrm{2}{t}−\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} \: \\ $$$${x}\left({t}\right)={e}^{\mathrm{2}{t}} \left(\frac{{e}^{\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} +{e}^{−\sqrt{\mathrm{2}}\boldsymbol{{i}}{t}} }{\mathrm{2}}\right) \\ $$$$\therefore{x}\left({t}\right)={e}^{\mathrm{2}{t}} \mathrm{cos}\left(\sqrt{\mathrm{2}}{t}\right) \\ $$$$\mathrm{Bromwich}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{defined}\:\mathrm{as} \\ $$$$\mathcal{L}_{{s}} ^{−\mathrm{1}} \left\{{f}\left({s}\right)\right\}=\:\frac{\mathrm{1}}{\mathrm{2}\pi\boldsymbol{{i}}}\:\oint_{\:\mathrm{C}} \:{f}\left({s}\right){e}^{{st}} \:\mathrm{d}{s} \\ $$$$\mathrm{if}\:\mathrm{complex}\:\mathrm{function}\:\:{f}\left({s}\right)\:\mathrm{is}\:\mathrm{entire} \\ $$$$\mathrm{Does}\:\mathcal{L}_{{s}} ^{−\mathrm{1}} \left\{{f}\left({s}\right)\right\}\:\mathrm{dosen}'\mathrm{t}\:\mathrm{Exist}? \\ $$$$\mathrm{for}\:\mathrm{example}.... \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left({r}\right){e}^{−{rt}} \mathrm{d}{r}=\frac{\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{and} \\ $$$$\mathrm{we}\:\mathrm{all}\:\mathrm{know}\:{J}_{\nu} \left({s}\right)=\oint_{\:\mathcal{C}} \:\:\frac{\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:{e}^{{st}} \:\mathrm{d}{t}\: \\ $$$$\mathrm{But}....\mathrm{can}'\mathrm{t}\:\mathrm{calculate}...\underset{{h}=\mathrm{1}} {\overset{{N}} {\sum}}\:\mathrm{Res}_{{t}={z}_{{h}} } \left\{\frac{\left({t}+\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}\right)^{−\nu} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\right\}\:{e}^{{st}} \\ $$$$\mathrm{and}\:{J}_{\nu} \left({s}\right)\:\mathrm{can}'\mathrm{t}\:\mathrm{express}\:\mathrm{by}\:{e}^{\lambda_{{h}} {t}} \:,\:{h}=\mathrm{1},\mathrm{2},\mathrm{3}... \\ $$$$\mathrm{and}\:\mathrm{by}\:\mathrm{my}\:\mathrm{Searching} \\ $$$${J}_{\nu} \left({s}\right)\:\mathrm{defined}\:\mathrm{as}\:\underset{{j}=\mathrm{0}} {\overset{\infty} {\sum}}\:\mathrm{Res}_{{t}=−{j}−\frac{\mathrm{1}}{\mathrm{2}}} \left\{\frac{\boldsymbol{\Gamma}\left({s}+\frac{\mathrm{1}}{\mathrm{2}}\nu\right)}{\boldsymbol{\Gamma}\left(\mathrm{1}−{s}+\frac{\mathrm{1}}{\mathrm{2}}\nu\right)}\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}{s}} \right\} \\ $$

Question Number 220660    Answers: 1   Comments: 0

Hmmm...... can you guys explain?? Let 𝛚=u dx+v dy be a 1-form defined over R^2 By applying the above formula to each terms consider x^1 =u , x^2 =v d𝛚=(Σ (∂u/∂x^i ) dx^i ∧dx)+(Σ (∂u/∂x^i ) dx^i ∧dy) =(∂u/∂x) dx∧dx+(∂u/∂y) dy∧dx+(∂u/∂x) dx∧dy+(∂u/∂y) dy∧dy 0−(∂u/∂y) dx∧dy+(∂u/∂x) dx∧dy+0 ((∂u/∂x)−(∂u/∂y)) dx∧dy dx∧dx=0 dx∧dy=−dy∧dx dy∧dy=0 i can′t understand why Exterior derivate ′′d𝛚=(Σ (∂u/∂x^i ) dx^i ∧dx)+(Σ (∂u/∂x^i ) dx^i ∧dy)′′ from the Stoke′s Theorem ∫_( ∂S) 𝛚=∫_( S) d𝛚 ∫_( ∂S) 𝛚=∫∫_( S) ((∂u/∂x)−(∂u/∂y)) dx∧dy

$$\mathrm{Hmmm}...... \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{explain}?? \\ $$$$\mathrm{Let}\:\boldsymbol{\omega}={u}\:\mathrm{d}{x}+{v}\:\mathrm{d}{y}\:\mathrm{be}\:\mathrm{a}\:\mathrm{1}-\mathrm{form}\:\mathrm{defined}\:\mathrm{over}\:\mathbb{R}^{\mathrm{2}} \\ $$$$\mathrm{By}\:\mathrm{applying}\:\mathrm{the}\:\mathrm{above}\:\mathrm{formula}\:\mathrm{to}\:\mathrm{each}\:\mathrm{terms} \\ $$$$\mathrm{consider}\:{x}^{\mathrm{1}} ={u}\:,\:{x}^{\mathrm{2}} ={v} \\ $$$$\mathrm{d}\boldsymbol{\omega}=\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{x}\right)+\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{y}\right) \\ $$$$=\frac{\partial{u}}{\partial{x}}\:\mathrm{d}{x}\wedge\mathrm{d}{x}+\frac{\partial{u}}{\partial{y}}\:\mathrm{d}{y}\wedge\mathrm{d}{x}+\frac{\partial{u}}{\partial{x}}\:\mathrm{d}{x}\wedge\mathrm{d}{y}+\frac{\partial{u}}{\partial{y}}\:\mathrm{d}{y}\wedge\mathrm{d}{y} \\ $$$$\mathrm{0}−\frac{\partial{u}}{\partial{y}}\:\mathrm{d}{x}\wedge\mathrm{d}{y}+\frac{\partial{u}}{\partial{x}}\:\mathrm{d}{x}\wedge\mathrm{d}{y}+\mathrm{0} \\ $$$$\left(\frac{\partial{u}}{\partial{x}}−\frac{\partial{u}}{\partial{y}}\right)\:\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$$$\mathrm{d}{x}\wedge\mathrm{d}{x}=\mathrm{0} \\ $$$$\mathrm{d}{x}\wedge\mathrm{d}{y}=−\mathrm{d}{y}\wedge\mathrm{d}{x} \\ $$$$\mathrm{d}{y}\wedge\mathrm{d}{y}=\mathrm{0} \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{understand}\:\mathrm{why}\:\mathrm{Exterior}\:\mathrm{derivate} \\ $$$$''\mathrm{d}\boldsymbol{\omega}=\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{\boldsymbol{{i}}} \wedge\mathrm{d}{x}\right)+\left(\Sigma\:\frac{\partial{u}}{\partial{x}^{{i}} }\:\mathrm{d}{x}^{{i}} \wedge\mathrm{d}{y}\right)'' \\ $$$$\mathrm{from}\:\mathrm{the}\:\mathrm{Stoke}'\mathrm{s}\:\mathrm{Theorem} \\ $$$$\int_{\:\partial{S}} \:\boldsymbol{\omega}=\int_{\:{S}} \:\mathrm{d}\boldsymbol{\omega} \\ $$$$\int_{\:\partial{S}} \:\boldsymbol{\omega}=\int\int_{\:{S}} \:\left(\frac{\partial{u}}{\partial{x}}−\frac{\partial{u}}{\partial{y}}\right)\:\mathrm{d}{x}\wedge\mathrm{d}{y} \\ $$

Question Number 220656    Answers: 3   Comments: 1

Question Number 220655    Answers: 0   Comments: 0

Question Number 220652    Answers: 0   Comments: 0

Show that ∫_0 ^( ∞) (e^(−st) /( (√(t^2 +1)))) dt=(1/2)π(H_0 ^ (s)−Y_0 (s)) , s∈R\{0}

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\frac{{e}^{−{st}} }{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{d}{t}=\frac{\mathrm{1}}{\mathrm{2}}\pi\left(\boldsymbol{\mathrm{H}}_{\mathrm{0}} ^{\:} \left({s}\right)−{Y}_{\mathrm{0}} \left({s}\right)\right)\:,\:{s}\in\mathbb{R}\backslash\left\{\mathrm{0}\right\} \\ $$

Question Number 220644    Answers: 2   Comments: 0

∫_1 ^( 2) ((2x^2 )/( (√((2x − 1)∙(2x + 2))))) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{1}} ^{\:\mathrm{2}} \:\frac{\mathrm{2}{x}^{\mathrm{2}} }{\:\sqrt{\left(\mathrm{2}{x}\:−\:\mathrm{1}\right)\centerdot\left(\mathrm{2}{x}\:+\:\mathrm{2}\right)}}\:{dx} \\ $$$$ \\ $$

Question Number 220642    Answers: 1   Comments: 0

Calculate i↑↑^∞ =?? i↑↑^∞ =i^i^i^⋰

$$\mathrm{Calculate} \\ $$$$\boldsymbol{{i}}\uparrow\uparrow^{\infty} =?? \\ $$$$\boldsymbol{{i}}\uparrow\uparrow^{\infty} =\boldsymbol{{i}}^{\boldsymbol{{i}}^{\boldsymbol{{i}}^{\iddots} } } \\ $$

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