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Question Number 204573    Answers: 0   Comments: 3

How Can derive LambertW(z) in the Form of integral??? W(z)=(1/π)∫_0 ^( π) ln(1+((z∙sin(t))/t)e^(t∙cot(t)) )dt , z∈[−(1/e),∞) Or Similar to the example.LambertW(z) How other Functions can be Derived in Integral Form

$$\mathrm{How}\:\mathrm{Can}\:\mathrm{derive}\:\mathrm{LambertW}\left({z}\right)\:\mathrm{in}\:\mathrm{the} \\ $$$$\:\mathrm{Form}\:\mathrm{of}\:\mathrm{integral}??? \\ $$$$\mathrm{W}\left({z}\right)=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{ln}\left(\mathrm{1}+\frac{{z}\centerdot\mathrm{sin}\left({t}\right)}{{t}}{e}^{{t}\centerdot\mathrm{cot}\left({t}\right)} \right)\mathrm{d}{t}\:,\:{z}\in\left[−\frac{\mathrm{1}}{{e}},\infty\right) \\ $$$$\mathrm{Or}\:\mathrm{Similar}\:\mathrm{to}\:\mathrm{the}\:\mathrm{example}.\mathrm{LambertW}\left({z}\right) \\ $$$$\mathrm{How}\:\mathrm{other}\:\mathrm{Functions}\:\mathrm{can}\:\mathrm{be}\:\mathrm{Derived}\:\mathrm{in}\:\mathrm{Integral}\:\mathrm{Form} \\ $$

Question Number 204569    Answers: 1   Comments: 0

find the value of I=∫_0 ^(+∞) ln(1+e^(−x) )dx nowing that Σ_(n=1) ^(+∞) (1/n^2 )=(π^2 /6)

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{value}}\:\boldsymbol{{of}}\: \\ $$$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{+\infty} \boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{e}}^{−\boldsymbol{{x}}} \right)\boldsymbol{{dx}}\:\boldsymbol{{nowing}}\:\boldsymbol{{that}}\: \\ $$$$\underset{{n}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$

Question Number 204533    Answers: 1   Comments: 0

Question Number 204522    Answers: 1   Comments: 3

Question Number 204517    Answers: 2   Comments: 0

Question Number 204472    Answers: 2   Comments: 0

Calculate ... Ω=Σ_(k=1) ^n ⌊(( 1)/( (e)^(1/k) −1)) ⌋ =?

$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Calculate}\:... \\ $$$$\:\:\:\:\:\:\:\Omega=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:\lfloor\frac{\:\mathrm{1}}{\:\sqrt[{{k}}]{{e}}\:−\mathrm{1}}\:\rfloor\:=? \\ $$$$ \\ $$

Question Number 204409    Answers: 3   Comments: 1

find ⌊∫_0 ^(2023) (2/(x+e^x ))dx⌋=?

$${find}\:\lfloor\int_{\mathrm{0}} ^{\mathrm{2023}} \frac{\mathrm{2}}{{x}+{e}^{{x}} }{dx}\rfloor=? \\ $$

Question Number 204275    Answers: 1   Comments: 0

Show that ∫_0 ^(π/4) (√(tan x)) (√(1−tan x)) dx=(((√((√2)−1))/( (√2)))−1)π

$$\mathrm{Show}\:\mathrm{that} \\ $$$$\underset{\mathrm{0}} {\overset{\frac{\pi}{\mathrm{4}}} {\int}}\sqrt{\mathrm{tan}\:{x}}\:\sqrt{\mathrm{1}−\mathrm{tan}\:{x}}\:{dx}=\left(\frac{\sqrt{\sqrt{\mathrm{2}}−\mathrm{1}}}{\:\sqrt{\mathrm{2}}}−\mathrm{1}\right)\pi \\ $$

Question Number 204244    Answers: 2   Comments: 0

Question Number 204233    Answers: 2   Comments: 0

Question Number 203964    Answers: 2   Comments: 0

Advanced calculus ... Q: If , ∫_0 ^1 ∫_0 ^( 1) (( xln(x)ln^2 (y ))/(1−xy)) dxdy = λ Σ_(n=1) ^∞ (( 1)/(n^( 3) ( n+1 )^2 )) ⇒ Find , λ =?

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{Advanced}\:\:{calculus}\:... \\ $$$$\:\:{Q}:\:\:\:{If}\:,\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{x}\mathrm{ln}\left({x}\right)\mathrm{ln}^{\mathrm{2}} \left({y}\:\right)}{\mathrm{1}−{xy}}\:{dxdy}\:=\:\lambda\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\mathrm{1}}{{n}^{\:\mathrm{3}} \left(\:{n}+\mathrm{1}\:\right)^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\:\:{Find}\:,\:\:\:\:\:\lambda\:=? \\ $$$$ \\ $$

Question Number 203884    Answers: 1   Comments: 0

find ∫_0 ^∞ (x^3 /((1+x)^4 (x+2)^5 ))dx

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}+{x}\right)^{\mathrm{4}} \left({x}+\mathrm{2}\right)^{\mathrm{5}} }{dx} \\ $$

Question Number 203867    Answers: 1   Comments: 0

find ∫(√((1−x^3 )/(1+x^3 )))dx

$${find}\:\int\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{3}} }{\mathrm{1}+{x}^{\mathrm{3}} }}{dx} \\ $$

Question Number 203836    Answers: 2   Comments: 0

Question Number 203772    Answers: 2   Comments: 0

Question Number 203747    Answers: 2   Comments: 0

Question Number 203679    Answers: 0   Comments: 0

Question Number 203564    Answers: 1   Comments: 0

∫_0 ^( ∞) ((sin^( 3) (x))/x^( 2) ) dx= ?

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{3}} \left({x}\right)}{{x}^{\:\mathrm{2}} }\:{dx}=\:?\:\:\:\:\: \\ $$

Question Number 203385    Answers: 2   Comments: 0

Question Number 203349    Answers: 1   Comments: 0

calculate ∫∫_([0,a]^2 ) e^(−x^2 −y^2 ) dxdy can you find ∫_0 ^a e^(−x^2 ) dx ? a>0

$${calculate}\:\int\int_{\left[\mathrm{0},{a}\right]^{\mathrm{2}} } \:{e}^{−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} } {dxdy} \\ $$$${can}\:{you}\:{find}\:\int_{\mathrm{0}} ^{{a}} {e}^{−{x}^{\mathrm{2}} } {dx}\:\:\:\:? \\ $$$${a}>\mathrm{0} \\ $$

Question Number 203186    Answers: 1   Comments: 0

f(x)={_(2 x=1) ^(7 x≠1 ) ⇒ ∫_0 ^( 4) f(x)dx=?

$${f}\left({x}\right)=\left\{_{\mathrm{2}\:\:\:\:\:\:\:\:{x}=\mathrm{1}} ^{\mathrm{7}\:\:\:\:\:\:\:\:{x}\neq\mathrm{1}\:\:\:\:\:} \Rightarrow\:\int_{\mathrm{0}} ^{\:\mathrm{4}} {f}\left({x}\right){dx}=?\right. \\ $$

Question Number 203047    Answers: 0   Comments: 0

Question Number 202930    Answers: 1   Comments: 0

Question Number 202882    Answers: 3   Comments: 0

∫_3 ^6 f(x)dx=5∫_1 ^6 f(x)dx=2 ∫_1 ^3 5f(x)dx

$$\:\:\: \underset{\mathrm{3}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{5}\underset{\mathrm{1}} {\overset{\mathrm{6}} {\int}}\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{2} \\ $$$$\:\:\: \\ $$$$\:\:\:\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\mathrm{5f}\left(\mathrm{x}\right)\mathrm{dx}\: \\ $$

Question Number 202636    Answers: 0   Comments: 0

((^3 (√4^(5−x) ))/(∫_4 ^6 (x−1)dx)) = (1/2^(2x−1) ) , find the value of x. Solution (4^((5−x)/3) /(∫_4 ^6 ((x^2 /2)−x+k))) = (1/2^(2x−1) ) (2^(2•((5−x)/3)) /(((6^2 /2)−6+k)−((4^2 /2)−4+k))) = (1/2^(2x−1) ) (2^((10−2x)/3) /(((36)/2)−6+k−((16)/2)+4−k)) = (1/2^(2x−1) ) (2^((10−2x)/3) /(18−6−8+4)) = (1/2^(2x−1) ) (2^((10−2x)/3) /8) = (1/2^(2x−1) ) (Cross Multiply) 2^(2x−1) ×2^((10−2x)/3) = 8×1 2^(2x−1) ×2^((10−2x)/3) = 2^3 2^(2x−1+((10−2x)/3)) = 2^3 (Since, the bases are equal. Then, we can equate the exponents) 2x−1+((10−2x)/3) = 3 (Multiply each term by 3) 3(2x)−3(1)+3(((10−2x)/3)) = 3(3) 6x−3+10−2x = 9 (Collect Like Terms) 4x+7 = 9 4x = 9−7 4x = 2 (Divide Both Sides by 4) ((4x)/4) = (2/4) ∴ x = (1/2) soln. by ibroclex_sunshine

$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\:^{\mathrm{3}} \sqrt{\mathrm{4}^{\mathrm{5}−\mathrm{x}} }}{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right){dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\mathrm{Solution}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}^{\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{x}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\mathrm{2}\bullet\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\left(\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{6}+\mathrm{k}\right)−\left(\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\frac{\mathrm{36}}{\mathrm{2}}−\mathrm{6}+\cancel{\mathrm{k}}−\frac{\mathrm{16}}{\mathrm{2}}+\mathrm{4}−\cancel{\mathrm{k}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{18}−\mathrm{6}−\mathrm{8}+\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:\:\left(\mathrm{Cross}\:\mathrm{Multiply}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{8}×\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \:\left(\mathrm{Since},\:\mathrm{the}\:\mathrm{bases}\:\mathrm{are}\:\mathrm{equal}.\:\mathrm{Then},\:\mathrm{we}\:\mathrm{can}\:\mathrm{equate}\:\mathrm{the}\:\mathrm{exponents}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}\:=\:\mathrm{3}\:\left(\mathrm{Multiply}\:\mathrm{each}\:\mathrm{term}\:\mathrm{by}\:\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{2x}\right)−\mathrm{3}\left(\mathrm{1}\right)+\cancel{\mathrm{3}}\left(\frac{\mathrm{10}−\mathrm{2x}}{\cancel{\mathrm{3}}}\right)\:=\:\mathrm{3}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6x}−\mathrm{3}+\mathrm{10}−\mathrm{2x}\:=\:\mathrm{9}\:\left(\mathrm{Collect}\:\mathrm{Like}\:\mathrm{Terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}+\mathrm{7}\:=\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{9}−\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{2}\:\left(\mathrm{Divide}\:\mathrm{Both}\:\mathrm{Sides}\:\mathrm{by}\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\cancel{\mathrm{4}x}}{\cancel{\mathrm{4}}}\:=\:\frac{\mathrm{2}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{soln}.\:\mathrm{by}\:\boldsymbol{\mathrm{ibroclex\_sunshine}}\: \\ $$$$ \\ $$

Question Number 202592    Answers: 0   Comments: 0

If I_n denotes ∫z^n e^(1/z) dz, then show that (n+1)!I_n =I_0 +e^(1/z) (1∙!z^2 +2∙!z^3 +∙∙∙+n!∙z^(n+1) )

$$\:\boldsymbol{{If}}\:\:\boldsymbol{{I}}_{\boldsymbol{{n}}} \:\boldsymbol{{denotes}}\:\int\boldsymbol{{z}}^{\boldsymbol{{n}}} \boldsymbol{{e}}^{\frac{\mathrm{1}}{\boldsymbol{{z}}}} \boldsymbol{{dz}},\:\boldsymbol{{then}}\:\boldsymbol{{show}}\:\boldsymbol{{that}} \\ $$$$\left(\boldsymbol{{n}}+\mathrm{1}\right)!\boldsymbol{{I}}_{\boldsymbol{{n}}} =\boldsymbol{{I}}_{\mathrm{0}} +\boldsymbol{{e}}^{\frac{\mathrm{1}}{\boldsymbol{{z}}}} \left(\mathrm{1}\centerdot!\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{2}\centerdot!\boldsymbol{{z}}^{\mathrm{3}} +\centerdot\centerdot\centerdot+\boldsymbol{{n}}!\centerdot\boldsymbol{{z}}^{\boldsymbol{{n}}+\mathrm{1}} \right) \\ $$$$ \\ $$

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