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Question Number 218629    Answers: 1   Comments: 0

Question Number 218626    Answers: 0   Comments: 1

Prove that for all real numbers a and b with a<b, the following inequality holds; (∫_a ^b 1 dx)^3 ≤ (b−a)(∫_a ^b (x−a+1)^2 dx)(∫_(a ) ^b (1/((a−x+1)^3 ))dx)

$$ \\ $$$$\:{Prove}\:{that}\:{for}\:{all}\:{real}\:{numbers}\:{a}\:{and}\:{b} \\ $$$${with}\:{a}<{b},\:{the}\:{following}\:{inequality}\:{holds}; \\ $$$$\left(\int_{{a}} ^{{b}} \mathrm{1}\:{dx}\right)^{\mathrm{3}} \leqslant\:\left({b}−{a}\right)\left(\int_{{a}} ^{{b}} \left({x}−{a}+\mathrm{1}\right)^{\mathrm{2}} {dx}\right)\left(\int_{{a}\:} ^{{b}} \frac{\mathrm{1}}{\left({a}−{x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}\right) \\ $$$$ \\ $$

Question Number 218624    Answers: 1   Comments: 0

Prove; ∫^e _(1/e) (t^2 /e^t^(2 ) ) dt ≤ 1− (1/e^(2 ) ) e−the base of natural logarithm

$$ \\ $$$$\:\:\:\:{Prove};\:\underset{\frac{\mathrm{1}}{{e}}} {\int}^{{e}} \:\frac{{t}^{\mathrm{2}} }{{e}^{{t}^{\mathrm{2}\:} } }\:{dt}\:\leqslant\:\mathrm{1}−\:\frac{\mathrm{1}}{{e}^{\mathrm{2}\:\:} } \\ $$$$\:\:\:{e}−{the}\:{base}\:{of}\:{natural}\:{logarithm} \\ $$$$ \\ $$

Question Number 219017    Answers: 0   Comments: 1

Question Number 218609    Answers: 1   Comments: 0

Can y=x be expressed as them su of two periodic functions?

$$\mathrm{Can}\:{y}={x}\:\mathrm{be}\:\mathrm{expressed}\:\mathrm{as}\:\mathrm{them} \\ $$$$\mathrm{su}\:\mathrm{of}\:\mathrm{two}\:\mathrm{periodic}\:\mathrm{functions}? \\ $$

Question Number 218607    Answers: 3   Comments: 0

Question Number 218606    Answers: 3   Comments: 0

Question Number 218605    Answers: 1   Comments: 0

Question Number 218603    Answers: 0   Comments: 3

Question Number 218599    Answers: 0   Comments: 2

∫_0 ^∞ x^(s−1) Π_(n=1 ) ^∞ (1−e^(−nx) )^(−24) dx

$$ \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\infty} \boldsymbol{{x}}^{\boldsymbol{{s}}−\mathrm{1}} \:\underset{\boldsymbol{{n}}=\mathrm{1}\:} {\overset{\infty} {\prod}}\left(\mathrm{1}−\boldsymbol{{e}}^{−\boldsymbol{{nx}}} \right)^{−\mathrm{24}} \:\boldsymbol{{dx}} \\ $$$$ \\ $$

Question Number 218598    Answers: 0   Comments: 0

∫^∞ _0 (x/(sinh(x)))ln((x/(e^x −1)))dx

$$ \\ $$$$\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{\boldsymbol{{x}}}{\boldsymbol{{sinh}}\left(\boldsymbol{{x}}\right)}\boldsymbol{{ln}}\left(\frac{\boldsymbol{{x}}}{\boldsymbol{{e}}^{\boldsymbol{{x}}} −\mathrm{1}}\right)\boldsymbol{{dx}}\: \\ $$$$ \\ $$

Question Number 218596    Answers: 0   Comments: 3

why my answer is delete???? Tinkutara why you ignore my effort

$$\mathrm{why}\:\mathrm{my}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{delete}???? \\ $$$$\mathrm{Tinkutara}\:\mathrm{why}\:\mathrm{you}\:\mathrm{ignore}\:\mathrm{my}\:\mathrm{effort} \\ $$

Question Number 218594    Answers: 0   Comments: 0

A kind of calculation relatedt o arctangent integral: Exere:∫_0 ^1 ((arctan^2 )/(x(1+x^2 )))dx. Solution:=∫_0 ^1 ((arctan^2 x)/x)dx_(A) −∫_0 ^1 ((x arctan^2 x)/(1+x^2 ))dx_(B) −∫_0 ^1 ((x arctan^2 )/((1+x^2 )^2 ))dx._(C) A=∫_0 ^1 ((arctan^2 x)/x)dx=∫_0 ^1 arctan^2 xd ln x=−2∫_0 ^1 ((ln x arctan x)/(1+x^2 ))dx =^(x→tan x) −2∫_0 ^(π/4) x ln tan xdx=^(Fourier series) −2∫_0 ^(π/4) x(−2Σ_(n=1) ^∞ ((cos(4n−2)x)/(2n−1)))dx =4Σ_(n=0) ^∞ (1/(2n−1))∫_0 ^(π/4) x cos(4n−2)xdx =4Σ_(n=1) ^∞ (1/(2n−1))∫_0 ^(π/4) x cos(4n−2)x dx =4Σ_(n=1) ^∞ (1/(2n−1)) ((−4+4cos(((4n−2)π)/4)+(4n−2)π sin(((4n−2)π)/4))/(4(4n−2)^2 )) =−Σ_(n=1) ^∞ (1/((2n−1)^3 ))+(π/2)Σ_(n=0) ^∞ (((−1)^(n−1) )/((2n−1)^2 ))=−(7/8)Σ_(n=1) ^∞ (1/n^3 )+(π/2)G = determinant (((−(7/8)ζ(3)+(π/2)G.)))(G Catalan) B=∫_0 ^1 ((x arctan^2 x)/(1+x^3 ))dx=^(x→tan x) ∫_0 ^(π/4) x^2 tan xdx=−∫_0 ^(π/4) x^2 d ln cos x =(π^2 /(32))ln 2+2∫_0 ^(π/4) x ln cos xdx =^(Fourier series) (π^2 /(32))ln 2+2∫_(0 ) ^(π/4) x[−ln 2+Σ_(n=1) ^∞ (((−1)^(n−1) )/n)cos 2nx]dx =−(π^2 /(32))ln 2−(7/(16))Σ_(n=0) ^∞ (((−1)^(n−1) )/n^3 )+(π/4)Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 )) =−(π^2 /(32))ln 2−((21)/(64))Σ_(n=1) ^∞ (1/n^3 )+(π/4)G =−(π^2 /(32))ln 2−((21)/(64))ζ(3)+(π/4)G C ∫((x arctan^2 x)/((1+x^2 )^2 ))dx=^(x→tan x) ∫_0 ^(π/4) x^2 sin x cos xdx=^(x→(π/2)) (1/(16))∫_0 ^(π/2) x^2 sin xdx =(1/(16))[2x sin x+(2−x^2 )cos x]_0 ^(π/2) =(π/(16))−(1/8) ∫_0 ^1 ((arctan^2 x)/(x(1+x^2 )^2 ))dx=A−B−C−(1/8)−(π/(16))+(π^2 /(64))ζ(3)+(π/4)G. Integral calculation of a kindof arctangent functionl reated to the above integral: (1)I(m,n)=∫_0 ^1 ((arctan^m x)/x^n )dx,(n≤m m,n∈N) (2)J(m,n)=∫_(0 ) ^(+∞) ((arctan^m x)/x^n )dx,(2≤n≤m m,n∈N) I(1,1)=∫_0 ^1 Σ_(n=1) ^∞ (((−1)^(n−1) x^(2n−2) )/(2n−1))dx=Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 ))=G. I(2,1)=−(7/8)ζ(3)+(π/2)G I(3,1)=∫_0 ^1 arctan^3 xd ln x=−3∫_0 ^1 ((arctan^2 x)/(1+x^2 ))ln xdx=^(x→tan x) −3∫_0 ^(π/4) x^2 ln tan xdx =^(Fourier series) −3∫_0 ^(π/4) x^2 (−2Σ_(n=1) ^∞ ((cos(4n−2)x)/(2n−1)))dx =(3/2)Σ_(n=1) ^∞ (((−1)^n )/((2n−1)^4 ))+((3π^2 )/(16))Σ_(n=1) ^∞ (((−1)^(n−1) )/((2n−1)^2 )) =(3/(512))Σ_(n=1) ^∞ (1/((n−(1/4))^4 ))−(3/(512))Σ_(n=1) ^∞ (1/((n−(1/4))^4 ))+((3π^2 )/(16))G =(1/(1024))[ψ^((3)) ((3/4))−ψ^((3)) ((1/4))]+((3π^2 )/(16))G. I(4,1)=(π/(1024))[ψ^((3)) −ψ^((3)) ((1/4))]+((93)/(32))ζ(5)+(π^3 /(16))G I(5,1)=((5π^2 )/(8192))[ψ^((3)) ((3/4))−ψ^((3)) ((1/4))]+(1/(65536))[ψ^((5)) ((1/4))−ψ^((5)) ((3/4))]+((5π^4 )/(256))G I(2,2)=∫_0 ^1 arctan^2 xd(1/x)−(π^2 /(16))+2∫_0 ^1 ((arctan x)/(x(1+x^2 )))dx =^(x→tan x) (π^2 /(16))+2∫_0 ^(π/4) x cot xdx=^(x→tan x) −(π^2 /(16))+2∫_0 ^(π/4) (ln 2+Σ_(n=1) ^∞ ((cos 2nx)/n))dx =−(π^2 /(16))+(π/4)ln 2+Σ_(n=0) ^∞ ((sin((nπ)/2))/n^2 )=−(π^2 /(16))+(π/4)ln 2+Σ_(n=1) ^∞ ((2(−1)^(n−1) )/((2n−1)^2 )) =−(π^2 /(16))+(π/4)ln 2+G I(3,2)=−(π^3 /(64))+((3π^2 )/(32))ln 2−((105)/(64))ζ(3)+((3π)/4)G. I(4,2)=(1/(512))[ψ^((3)) ((3/4))−ψ^((3)) ((1/4))]=(π^4 /(256))+(π^3 /(32))ln 2−((9π)/(64))ζ(3)+((3π^3 )/8)G. I(3,3)=−((3π^2 )/(32))−(π^3 /(64))+((9π)/8)ln 2−((105)/(32))ζ(3)+((3π)/2)G. It is evident that I(m,n)has elegant closedforms value composed of ζ(2k+1), ψ^((2k+1)) ((3/4))−ψ^((2k+1)) ((1/4)),π^k , Catalan constant G(for k=1,2,3,…) J(2,2)=π ln 2. J(3,2)=((3π^2 )/4)ln 2−((81)/8)ζ(3). J(4,2)=(π^3 /2)ln 2−((9π)/4)ζ(3) J(5,2)=((5π^4 )/(16))ln 2−((45π^2 )/(16))ζ(3)+((465)/(32))ζ(5). J(3,3)=−(π^3 /(16))+((3π)/2)ln 2. J(4,3)=−(π^4 /(32))+((3π^2 )/2)ln 2−((21)/4)ζ(3) J(5,3)=(π^6 /(64))+((5π^3 )/4)ln 2−((45π)/8)ζ(3). J(4,4)=−(π^3 /(12))+(2π−(π^3 /6))ln 2+((3π)/4)ζ(3) J(5,4)=−((5π^4 )/(96))+(((5π^2 )/2)−((5π^4 )/(48)))ln 2−(((35)/4)+((15π^2 )/(16)))ζ(3)−((155)/(32))ζ(5). J(5,5)=(π^5 /(128))−((5π^3 )/(48))+(((5π)/2)−((5π^3 )/6))ln 2+((15π)/4)ζ(3). It is evident that J(m,n)also has elegant closed-form values composed of ζ(2k+1),π^k (for k=1,2,3,…).

$$\mathrm{A}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{calculation}\:\mathrm{relatedt} \\ $$$$\mathrm{o}\:\mathrm{arctangent}\:\mathrm{integral}: \\ $$$$\boldsymbol{\mathrm{Exere}}:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} }{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}. \\ $$$$\mathrm{Solution}:\underset{{A}} {\underbrace{=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} {x}}{{x}}{dx}}}\:−\underset{{B}} {\underbrace{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\:\mathrm{arctan}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}}}\:−\underset{{C}} {\underbrace{\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\:\mathrm{arctan}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}.}} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} {x}}{{x}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}^{\mathrm{2}} {xd}\:\mathrm{ln}\:{x}=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\:{x}\:\mathrm{arctan}\:{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\overset{{x}\rightarrow\mathrm{tan}\:{x}} {=}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\:\mathrm{ln}\:\mathrm{tan}\:{xdx}\overset{\boldsymbol{\mathrm{Fourier}}\:\boldsymbol{\mathrm{series}}} {=}−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\left(−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\left(\mathrm{4}{n}−\mathrm{2}\right){x}}{\mathrm{2}{n}−\mathrm{1}}\right){dx} \\ $$$$=\mathrm{4}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\:\mathrm{cos}\left(\mathrm{4}{n}−\mathrm{2}\right){xdx} \\ $$$$=\mathrm{4}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\:\mathrm{cos}\left(\mathrm{4}{n}−\mathrm{2}\right){x}\:{dx} \\ $$$$=\mathrm{4}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:\frac{−\mathrm{4}+\mathrm{4cos}\frac{\left(\mathrm{4}{n}−\mathrm{2}\right)\pi}{\mathrm{4}}+\left(\mathrm{4}{n}−\mathrm{2}\right)\pi\:\mathrm{sin}\frac{\left(\mathrm{4}{n}−\mathrm{2}\right)\pi}{\mathrm{4}}}{\mathrm{4}\left(\mathrm{4}{n}−\mathrm{2}\right)^{\mathrm{2}} } \\ $$$$=−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{7}}{\mathrm{8}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }+\frac{\pi}{\mathrm{2}}\boldsymbol{\mathrm{G}} \\ $$$$=\begin{array}{|c|}{−\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\frac{\pi}{\mathrm{2}}\boldsymbol{\mathrm{G}}.}\\\hline\end{array}\left(\boldsymbol{\mathrm{G}}\:\mathrm{Catalan}\right) \\ $$$$\boldsymbol{{B}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}\:\mathrm{arctan}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\overset{{x}\rightarrow\mathrm{tan}\:{x}} {=}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{2}} \mathrm{tan}\:{xdx}=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{2}} {d}\:\mathrm{ln}\:\mathrm{cos}\:{x} \\ $$$$=\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\:\mathrm{ln}\:\mathrm{cos}\:{xdx} \\ $$$$\overset{\boldsymbol{\mathrm{Fourier}}\:\boldsymbol{\mathrm{series}}} {=}\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}+\mathrm{2}\int_{\mathrm{0}\:} ^{\frac{\pi}{\mathrm{4}}} {x}\left[−\mathrm{ln}\:\mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\mathrm{cos}\:\mathrm{2}{nx}\right]{dx} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{7}}{\mathrm{16}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} }+\frac{\pi}{\mathrm{4}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{21}}{\mathrm{64}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{3}} }+\frac{\pi}{\mathrm{4}}\boldsymbol{\mathrm{G}} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{21}}{\mathrm{64}}\zeta\left(\mathrm{3}\right)+\frac{\pi}{\mathrm{4}}\boldsymbol{\mathrm{G}} \\ $$$${C}\:\:\:\:\:\int\frac{{x}\:\mathrm{arctan}^{\mathrm{2}} {x}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\overset{{x}\rightarrow\mathrm{tan}\:{x}} {=}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{2}} \mathrm{sin}\:{x}\:\mathrm{cos}\:{xdx}\overset{{x}\rightarrow\frac{\pi}{\mathrm{2}}} {=}\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {x}^{\mathrm{2}} \mathrm{sin}\:{xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\left[\mathrm{2}{x}\:\mathrm{sin}\:{x}+\left(\mathrm{2}−{x}^{\mathrm{2}} \right)\mathrm{cos}\:{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi}{\mathrm{16}}−\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} {x}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}={A}−{B}−{C}−\frac{\mathrm{1}}{\mathrm{8}}−\frac{\pi}{\mathrm{16}}+\frac{\pi^{\mathrm{2}} }{\mathrm{64}}\zeta\left(\mathrm{3}\right)+\frac{\pi}{\mathrm{4}}\boldsymbol{\mathrm{G}}. \\ $$$$\mathrm{Integral}\:\mathrm{calculation}\:\mathrm{of}\:\mathrm{a}\: \\ $$$$\mathrm{kindof}\:\mathrm{arctangent}\:\mathrm{functionl} \\ $$$$\mathrm{reated}\:\mathrm{to}\:\mathrm{the}\:\mathrm{above}\:\mathrm{integral}: \\ $$$$\left(\mathrm{1}\right){I}\left({m},{n}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{{m}} {x}}{{x}^{{n}} }{dx},\left({n}\leqslant{m}\:{m},{n}\in\mathbb{N}\right) \\ $$$$\left(\mathrm{2}\right){J}\left({m},{n}\right)=\int_{\mathrm{0}\:} ^{+\infty} \frac{\mathrm{arctan}^{{m}} {x}}{{x}^{{n}} }{dx},\left(\mathrm{2}\leqslant{n}\leqslant{m}\:{m},{n}\in\mathbb{N}\right) \\ $$$${I}\left(\mathrm{1},\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {x}^{\mathrm{2}{n}−\mathrm{2}} }{\mathrm{2}{n}−\mathrm{1}}{dx}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }=\boldsymbol{\mathrm{G}}. \\ $$$${I}\left(\mathrm{2},\mathrm{1}\right)=−\frac{\mathrm{7}}{\mathrm{8}}\zeta\left(\mathrm{3}\right)+\frac{\pi}{\mathrm{2}}\boldsymbol{\mathrm{G}} \\ $$$${I}\left(\mathrm{3},\mathrm{1}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}^{\mathrm{3}} {xd}\:\mathrm{ln}\:{x}=−\mathrm{3}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}^{\mathrm{2}} {x}}{\mathrm{1}+{x}^{\mathrm{2}} }\mathrm{ln}\:{xdx}\overset{{x}\rightarrow\mathrm{tan}\:{x}} {=}−\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{2}} \mathrm{ln}\:\mathrm{tan}\:{xdx} \\ $$$$\overset{\boldsymbol{\mathrm{Fourier}}\:\boldsymbol{\mathrm{series}}} {=}−\mathrm{3}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}^{\mathrm{2}} \left(−\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\left(\mathrm{4}{n}−\mathrm{2}\right){x}}{\mathrm{2}{n}−\mathrm{1}}\right){dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{4}} }+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{16}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{3}}{\mathrm{512}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} }−\frac{\mathrm{3}}{\mathrm{512}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left({n}−\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} }+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{16}}\boldsymbol{\mathrm{G}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1024}}\left[\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right]+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{16}}\boldsymbol{\mathrm{G}}. \\ $$$${I}\left(\mathrm{4},\mathrm{1}\right)=\frac{\pi}{\mathrm{1024}}\left[\psi^{\left(\mathrm{3}\right)} −\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right]+\frac{\mathrm{93}}{\mathrm{32}}\zeta\left(\mathrm{5}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{16}}\boldsymbol{\mathrm{G}} \\ $$$${I}\left(\mathrm{5},\mathrm{1}\right)=\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{8192}}\left[\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right]+\frac{\mathrm{1}}{\mathrm{65536}}\left[\psi^{\left(\mathrm{5}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)−\psi^{\left(\mathrm{5}\right)} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)\right]+\frac{\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{256}}\boldsymbol{\mathrm{G}} \\ $$$${I}\left(\mathrm{2},\mathrm{2}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{arctan}^{\mathrm{2}} {xd}\frac{\mathrm{1}}{{x}}−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{arctan}\:{x}}{{x}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\ $$$$\overset{{x}\rightarrow\mathrm{tan}\:{x}} {=}\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {x}\:\mathrm{cot}\:{xdx}\overset{{x}\rightarrow\mathrm{tan}\:{x}} {=}−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{ln}\:\mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{cos}\:\mathrm{2}{nx}}{{n}}\right){dx} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}+\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\frac{{n}\pi}{\mathrm{2}}}{{n}^{\mathrm{2}} }=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}+\frac{\pi}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}+\boldsymbol{\mathrm{G}} \\ $$$${I}\left(\mathrm{3},\mathrm{2}\right)=−\frac{\pi^{\mathrm{3}} }{\mathrm{64}}+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{105}}{\mathrm{64}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{3}\pi}{\mathrm{4}}\boldsymbol{\mathrm{G}}. \\ $$$${I}\left(\mathrm{4},\mathrm{2}\right)=\frac{\mathrm{1}}{\mathrm{512}}\left[\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi^{\left(\mathrm{3}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\right]=\frac{\pi^{\mathrm{4}} }{\mathrm{256}}+\frac{\pi^{\mathrm{3}} }{\mathrm{32}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{9}\pi}{\mathrm{64}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{3}\pi^{\mathrm{3}} }{\mathrm{8}}\boldsymbol{\mathrm{G}}. \\ $$$${I}\left(\mathrm{3},\mathrm{3}\right)=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{32}}−\frac{\pi^{\mathrm{3}} }{\mathrm{64}}+\frac{\mathrm{9}\pi}{\mathrm{8}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{105}}{\mathrm{32}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{3}\pi}{\mathrm{2}}\boldsymbol{\mathrm{G}}. \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{evident}\:\mathrm{that}\:{I}\left({m},{n}\right)\mathrm{has}\:\mathrm{elegant}\:\mathrm{closedforms} \\ $$$$\mathrm{value}\:\mathrm{composed}\:\mathrm{of}\:\zeta\left(\mathrm{2}{k}+\mathrm{1}\right), \\ $$$$\psi^{\left(\mathrm{2}{k}+\mathrm{1}\right)} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)−\psi^{\left(\mathrm{2}{k}+\mathrm{1}\right)} \left(\frac{\mathrm{1}}{\mathrm{4}}\right),\pi^{{k}} , \\ $$$$\mathrm{Catalan}\:\mathrm{constant}\:{G}\left(\mathrm{for}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3},\ldots\right) \\ $$$${J}\left(\mathrm{2},\mathrm{2}\right)=\pi\:\mathrm{ln}\:\mathrm{2}. \\ $$$${J}\left(\mathrm{3},\mathrm{2}\right)=\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{81}}{\mathrm{8}}\zeta\left(\mathrm{3}\right). \\ $$$${J}\left(\mathrm{4},\mathrm{2}\right)=\frac{\pi^{\mathrm{3}} }{\mathrm{2}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{9}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$${J}\left(\mathrm{5},\mathrm{2}\right)=\frac{\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{16}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{45}\pi^{\mathrm{2}} }{\mathrm{16}}\zeta\left(\mathrm{3}\right)+\frac{\mathrm{465}}{\mathrm{32}}\zeta\left(\mathrm{5}\right). \\ $$$${J}\left(\mathrm{3},\mathrm{3}\right)=−\frac{\pi^{\mathrm{3}} }{\mathrm{16}}+\frac{\mathrm{3}\pi}{\mathrm{2}}\mathrm{ln}\:\mathrm{2}. \\ $$$${J}\left(\mathrm{4},\mathrm{3}\right)=−\frac{\pi^{\mathrm{4}} }{\mathrm{32}}+\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{2}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{21}}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$${J}\left(\mathrm{5},\mathrm{3}\right)=\frac{\pi^{\mathrm{6}} }{\mathrm{64}}+\frac{\mathrm{5}\pi^{\mathrm{3}} }{\mathrm{4}}\mathrm{ln}\:\mathrm{2}−\frac{\mathrm{45}\pi}{\mathrm{8}}\zeta\left(\mathrm{3}\right). \\ $$$${J}\left(\mathrm{4},\mathrm{4}\right)=−\frac{\pi^{\mathrm{3}} }{\mathrm{12}}+\left(\mathrm{2}\pi−\frac{\pi^{\mathrm{3}} }{\mathrm{6}}\right)\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{3}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right) \\ $$$${J}\left(\mathrm{5},\mathrm{4}\right)=−\frac{\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{96}}+\left(\frac{\mathrm{5}\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{5}\pi^{\mathrm{4}} }{\mathrm{48}}\right)\mathrm{ln}\:\mathrm{2}−\left(\frac{\mathrm{35}}{\mathrm{4}}+\frac{\mathrm{15}\pi^{\mathrm{2}} }{\mathrm{16}}\right)\zeta\left(\mathrm{3}\right)−\frac{\mathrm{155}}{\mathrm{32}}\zeta\left(\mathrm{5}\right). \\ $$$${J}\left(\mathrm{5},\mathrm{5}\right)=\frac{\pi^{\mathrm{5}} }{\mathrm{128}}−\frac{\mathrm{5}\pi^{\mathrm{3}} }{\mathrm{48}}+\left(\frac{\mathrm{5}\pi}{\mathrm{2}}−\frac{\mathrm{5}\pi^{\mathrm{3}} }{\mathrm{6}}\right)\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{15}\pi}{\mathrm{4}}\zeta\left(\mathrm{3}\right). \\ $$$$\mathrm{It}\:\mathrm{is}\:\mathrm{evident}\:\mathrm{that}\:{J}\left({m},{n}\right)\mathrm{also}\:\mathrm{has}\: \\ $$$$\mathrm{elegant}\:\mathrm{closed}-\mathrm{form}\:\mathrm{values}\:\mathrm{composed}\:\mathrm{of}\:\zeta\left(\mathrm{2}{k}+\mathrm{1}\right),\pi^{{k}} \left(\mathrm{for}\:{k}=\mathrm{1},\mathrm{2},\mathrm{3},\ldots\right). \\ $$

Question Number 218583    Answers: 1   Comments: 0

∫_(0 ) ^∞ e^(−x) (Σ_(n=1) ^∞ ((f(n))/n) sin(nx))dx =1

$$ \\ $$$$\:\int_{\mathrm{0}\:} ^{\infty} \:{e}^{−{x}} \:\left(\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{f}\left({n}\right)}{{n}}\:{sin}\left({nx}\right)\right){dx}\:=\mathrm{1} \\ $$$$ \\ $$

Question Number 218582    Answers: 1   Comments: 0

∫_0 ^∞ ((cos(ax) − cos(bx))/x^2 ) dx = (π/2) ∣b−a∣

$$ \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{cos}\left({ax}\right)\:−\:{cos}\left({bx}\right)}{{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{\pi}{\mathrm{2}}\:\mid{b}−{a}\mid\: \\ $$$$ \\ $$

Question Number 218580    Answers: 1   Comments: 3

Question Number 218578    Answers: 2   Comments: 0

10 couples are invited to a dinner. after the dinner they form pairs to dance. in how many ways can they do this, 1) generally 2) a man should dance with a woman 3) as 2), but two special couples should not dance with each other 4) a man should dance with a woman, but not with his own wife 5) as 4), but two special couples should not dance with each other

$$\mathrm{10}\:{couples}\:{are}\:{invited}\:{to}\:{a}\:{dinner}. \\ $$$${after}\:{the}\:{dinner}\:{they}\:{form}\:{pairs}\: \\ $$$${to}\:{dance}.\:{in}\:{how}\:{many}\:{ways}\:{can}\:{they} \\ $$$${do}\:{this}, \\ $$$$\left.\mathrm{1}\right)\:{generally} \\ $$$$\left.\mathrm{2}\right)\:{a}\:{man}\:{should}\:{dance}\:{with}\:{a}\:{woman} \\ $$$$\left.\mathrm{3}\left.\right)\:{as}\:\mathrm{2}\right),\:{but}\:{two}\:{special}\:{couples} \\ $$$$\:\:\:\:\:{should}\:{not}\:{dance}\:{with}\:{each}\:{other} \\ $$$$\left.\mathrm{4}\right)\:{a}\:{man}\:{should}\:{dance}\:{with}\:{a}\:{woman}, \\ $$$$\:\:\:\:\:{but}\:{not}\:{with}\:{his}\:{own}\:{wife} \\ $$$$\left.\mathrm{5}\left.\right)\:{as}\:\mathrm{4}\right),\:{but}\:{two}\:{special}\:{couples} \\ $$$$\:\:\:\:\:{should}\:{not}\:{dance}\:{with}\:{each}\:{other} \\ $$

Question Number 218563    Answers: 0   Comments: 0

solve for x ∈ R ∫_0 ^∞ ((sin(xt))/(e^t −1 )) dt = (𝛑/2) coth(𝛑x) − (1/(2x))

$$ \\ $$$$\:\:\:\:\:\:\:\:\:{solve}\:{for}\:\boldsymbol{{x}}\:\in\:\mathbb{R}\: \\ $$$$\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sin}\left({xt}\right)}{{e}^{{t}} −\mathrm{1}\:}\:{dt}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{2}}\:{coth}\left(\boldsymbol{\pi}{x}\right)\:−\:\frac{\mathrm{1}}{\mathrm{2}{x}}\:\: \\ $$$$ \\ $$

Question Number 218560    Answers: 0   Comments: 1

Question Number 218557    Answers: 1   Comments: 1

$$\: \\ $$

Question Number 218547    Answers: 1   Comments: 1

Question Number 218543    Answers: 1   Comments: 1

Question Number 218539    Answers: 1   Comments: 0

Solve (∂^2 w/∂t^2 )=c^2 (∂^2 w/∂x^2 ) w(0,t)=f(t) ,lim_(x→∞) w(x,t)=0 (Boundary Condition) w(x,0)=0 , w_t (x,0)=0 (Initial Condition) f(t) { ((sin(t) , t∈[0,2π))),((0 , otherwise)) :}

$${S}\mathrm{olve} \\ $$$$\frac{\partial^{\mathrm{2}} {w}}{\partial{t}^{\mathrm{2}} }={c}^{\mathrm{2}} \frac{\partial^{\mathrm{2}} {w}}{\partial{x}^{\mathrm{2}} } \\ $$$${w}\left(\mathrm{0},{t}\right)={f}\left({t}\right)\:,\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{w}\left({x},{t}\right)=\mathrm{0}\:\left(\mathrm{Boundary}\:\mathrm{Condition}\right) \\ $$$${w}\left({x},\mathrm{0}\right)=\mathrm{0}\:,\:{w}_{{t}} \left({x},\mathrm{0}\right)=\mathrm{0}\:\left(\mathrm{Initial}\:\mathrm{Condition}\right) \\ $$$${f}\left({t}\right)\begin{cases}{\mathrm{sin}\left({t}\right)\:,\:{t}\in\left[\mathrm{0},\mathrm{2}\pi\right)}\\{\mathrm{0}\:,\:\mathrm{otherwise}}\end{cases} \\ $$

Question Number 218532    Answers: 0   Comments: 0

Question Number 218534    Answers: 0   Comments: 0

I Find Fun integral problem! ∫ x^dx −1 =?? note) I already know that integral Solution Try integral problem! (# Product Integral , #Integral)

$$\mathrm{I}\:\mathrm{Find}\:\mathrm{Fun}\:\mathrm{integral}\:\mathrm{problem}! \\ $$$$\int\:\:{x}^{\mathrm{d}{x}} −\mathrm{1}\:=??\: \\ $$$$\left.\mathrm{note}\right)\:\:\mathrm{I}\:\mathrm{already}\:\mathrm{know}\:\mathrm{that}\:\mathrm{integral}\:\mathrm{Solution} \\ $$$$\:\mathrm{Try}\:\mathrm{integral}\:\mathrm{problem}! \\ $$$$\left(#\:\mathrm{Product}\:\mathrm{Integral}\:,\:#\mathrm{Integral}\right) \\ $$

Question Number 218528    Answers: 3   Comments: 0

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