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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^⋰

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

Question Number 220630    Answers: 1   Comments: 0

Question Number 220628    Answers: 0   Comments: 0

Question Number 220627    Answers: 3   Comments: 0

Question Number 220626    Answers: 2   Comments: 0

Question Number 220625    Answers: 1   Comments: 0

Question Number 220624    Answers: 0   Comments: 0

Question Number 220602    Answers: 2   Comments: 0

Question Number 220607    Answers: 1   Comments: 1

symplify (((√3)+(√(5+))(√9)+(√(15)))/( (√1)+(√5)+(√(12))))

$${symplify} \\ $$$$\frac{\sqrt{\mathrm{3}}+\sqrt{\mathrm{5}+}\sqrt{\mathrm{9}}+\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{1}}+\sqrt{\mathrm{5}}+\sqrt{\mathrm{12}}} \\ $$

Question Number 220606    Answers: 1   Comments: 1

Question Number 220590    Answers: 1   Comments: 0

∫_0 ^( ∞) wCi(w)e^(−w) dw=?? Ci(w)=−∫_w ^( ∞) ((cos(t))/t) dt

$$\int_{\mathrm{0}} ^{\:\infty} \:\:{w}\mathrm{Ci}\left({w}\right){e}^{−{w}} \:\mathrm{d}{w}=?? \\ $$$$\mathrm{Ci}\left({w}\right)=−\int_{{w}} ^{\:\infty} \:\:\frac{\mathrm{cos}\left({t}\right)}{{t}}\:\mathrm{d}{t} \\ $$

Question Number 220588    Answers: 0   Comments: 1

Eucleadian Space R^2 and Subset A A={(x,y)∈R^2 ∣x^2 +y^2 =1}, B={(((t−1)/t) cos(t),((t−1)/t)sin(t))∈R^2 ∣1≤t∈R} Show that X=A∪B is Connect set

$$\mathrm{Eucleadian}\:\mathrm{Space}\:\mathbb{R}^{\mathrm{2}} \:\mathrm{and}\:\mathrm{Subset}\:{A} \\ $$$${A}=\left\{\left({x},{y}\right)\in\mathbb{R}^{\mathrm{2}} \mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{1}\right\},\:{B}=\left\{\left(\frac{{t}−\mathrm{1}}{{t}}\:\mathrm{cos}\left({t}\right),\frac{{t}−\mathrm{1}}{{t}}\mathrm{sin}\left({t}\right)\right)\in\mathbb{R}^{\mathrm{2}} \mid\mathrm{1}\leq{t}\in\mathbb{R}\right\} \\ $$$$\mathrm{Show}\:\mathrm{that}\:{X}={A}\cup{B}\:\mathrm{is}\:\mathrm{Connect}\:\mathrm{set} \\ $$

Question Number 220583    Answers: 0   Comments: 0

Question Number 220581    Answers: 1   Comments: 0

An object is thrown vertically upward from the top of a building 20m high. If the object passes the point it was thrown 4 seconds on it way down, find the. (a) Velocity at which the object was thrown. (b) time taken when the object is 10m above the level it was thrown. (c) time taken when the object is 10m below the level it was thrown. (d) Velocity with which the object hit the ground. [Take g = 10m/s²]

An object is thrown vertically upward from the top of a building 20m high. If the object passes the point it was thrown 4 seconds on it way down, find the. (a) Velocity at which the object was thrown. (b) time taken when the object is 10m above the level it was thrown. (c) time taken when the object is 10m below the level it was thrown. (d) Velocity with which the object hit the ground. [Take g = 10m/s²]

Question Number 220580    Answers: 0   Comments: 0

∫_(−∞i) ^(+∞i) ((atan(w))/w)e^(3iw) dw=??

$$\int_{−\infty\boldsymbol{{i}}} ^{+\infty\boldsymbol{{i}}} \:\frac{\mathrm{atan}\left({w}\right)}{{w}}{e}^{\mathrm{3}\boldsymbol{{i}}{w}} \mathrm{d}{w}=?? \\ $$

Question Number 220577    Answers: 1   Comments: 0

Calculate the perimeter of a rectangle whose area is represented by the polynomial 25x^2 −35x+12(Given that the length and breath are not constant)

$${Calculate}\:{the}\:{perimeter}\:{of}\:{a}\:{rectangle} \\ $$$${whose}\:{area}\:{is}\:{represented}\:{by}\:{the}\:{polynomial} \\ $$$$\mathrm{25}{x}^{\mathrm{2}} −\mathrm{35}{x}+\mathrm{12}\left({Given}\:{that}\:{the}\:{length}\:{and}\:{breath}\:{are}\:{not}\:{constant}\right) \\ $$

Question Number 220579    Answers: 1   Comments: 0

A=7×19×31×43×.....upto 29 terms find the last four digits of A.

$$\:{A}=\mathrm{7}×\mathrm{19}×\mathrm{31}×\mathrm{43}×.....{upto}\:\mathrm{29}\:{terms} \\ $$$$\:{find}\:{the}\:{last}\:{four}\:{digits}\:{of}\:{A}. \\ $$

Question Number 220563    Answers: 1   Comments: 0

Calculate the exact value of : I=∫_0 ^4 e^(−x^2 ) dx

$${Calculate}\:{the}\:{exact}\:{value}\:{of}\:: \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{4}} {e}^{−{x}^{\mathrm{2}} } {dx} \\ $$

Question Number 220562    Answers: 1   Comments: 0

∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1)) dz=I I(t)=∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((z∙ln(z^2 +1))/(z^2 +1))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((z^2 ln(z^2 +1)+ln(z^2 +1)−ln(z^2 +1))/(z(z^2 +1)))e^(−zt) dz I′(t)=−∫_0 ^( ∞) ((ln(z^2 +1))/z)e^(−zt ) dz+∫_0 ^( ∞) ((ln(z^2 +1))/(z(z^2 +1)))e^(−zt) dz I′′(t)=∫_0 ^∞ ln(z^2 +1)e^(−zt) dz−∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1))e^(−zt) dz I^((2)) (t)+I(t)=∫_0 ^( ∞) ln(z^2 +1)e^(−zt) dz........(A) ∫_0 ^( ∞) ln(z^2 +1)e^(−zt) dz cos^2 (α)+sin^2 (α)=1 → 1+tan^2 (α)=sec^2 (α) z=tan(u) → u=tan^(−1) (z) (dz/du)=sec^2 (u) → dz=sec^2 (u)du ∫_0 ^( ∞) ln(z^2 +1)e^(−st) dz=∫_0 ^( (π/2)) sec^2 (u)∙ln(sec^2 (u))e^(−s∙tan(u)) du i can′t Calculate anymore.... can′t solve ODE....(A) that Equation(A) Seems to Weird cus Feynman trick ...... How can i solve ∫_0 ^( ∞) ((ln(z^2 +1))/(z^2 +1)) dz or....Should i do Complex integral.. ∮_( C) f(z) dz.....??

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{z}\centerdot\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{{z}^{\mathrm{2}} \mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)+\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}'\left({t}\right)=−\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}}{e}^{−{zt}\:} \mathrm{d}{z}+\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}''\left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}−\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}{e}^{−{zt}} \:\mathrm{d}{z} \\ $$$${I}^{\left(\mathrm{2}\right)} \left({t}\right)+{I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z}........\left(\mathrm{A}\right) \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{zt}} \mathrm{d}{z} \\ $$$$\mathrm{cos}^{\mathrm{2}} \left(\alpha\right)+\mathrm{sin}^{\mathrm{2}} \left(\alpha\right)=\mathrm{1}\:\rightarrow\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \left(\alpha\right)=\mathrm{sec}^{\mathrm{2}} \left(\alpha\right) \\ $$$${z}=\mathrm{tan}\left({u}\right)\:\rightarrow\:{u}=\mathrm{tan}^{−\mathrm{1}} \left({z}\right) \\ $$$$\frac{\mathrm{d}{z}}{\mathrm{d}{u}}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\:\rightarrow\:\mathrm{d}{z}=\mathrm{sec}^{\mathrm{2}} \left({u}\right)\mathrm{d}{u} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right){e}^{−{st}} \mathrm{d}{z}=\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\mathrm{sec}^{\mathrm{2}} \left({u}\right)\centerdot\mathrm{ln}\left(\mathrm{sec}^{\mathrm{2}} \left({u}\right)\right){e}^{−{s}\centerdot\mathrm{tan}\left({u}\right)} \mathrm{d}{u} \\ $$$$\mathrm{i}\:\mathrm{can}'\mathrm{t}\:\mathrm{Calculate}\:\mathrm{anymore}.... \\ $$$$\mathrm{can}'\mathrm{t}\:\mathrm{solve}\:\mathrm{ODE}....\left(\mathrm{A}\right) \\ $$$$\mathrm{that}\:\mathrm{Equation}\left(\mathrm{A}\right)\:\mathrm{Seems}\:\mathrm{to}\:\mathrm{Weird}\:\mathrm{cus}\:\mathrm{Feynman}\:\mathrm{trick}\: \\ $$$$...... \\ $$$$\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{solve}\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{ln}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}{z} \\ $$$$\mathrm{or}....\mathrm{Should}\:\mathrm{i}\:\mathrm{do}\:\mathrm{Complex}\:\mathrm{integral}.. \\ $$$$\oint_{\:{C}} \:{f}\left({z}\right)\:\mathrm{d}{z}.....?? \\ $$

Question Number 220560    Answers: 3   Comments: 0

sinα=0.8 ⇒ ((BE)/(EF))=?

$${sin}\alpha=\mathrm{0}.\mathrm{8}\:\Rightarrow\:\frac{{BE}}{{EF}}=? \\ $$

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