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Prove; ∫_0 ^( 1) ((((√(x )) ln (√x))^2 )/((1−x^2 )^2 + 4x^2 + 2x^4 + 3x^6 + 2x^8 + x^(10) )) dx = ((21)/(1024)) ζ(3) − (π^3 /(1024)) + (π^3 /(324(√3)))

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove}; \\ $$$$\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\left(\sqrt{{x}\:}\:\mathrm{ln}\:\sqrt{{x}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:+\:\mathrm{4}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}^{\mathrm{4}} \:+\:\mathrm{3}{x}^{\mathrm{6}} \:+\:\mathrm{2}{x}^{\mathrm{8}} \:+\:{x}^{\mathrm{10}} }\:\:\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:=\:\frac{\mathrm{21}}{\mathrm{1024}}\:\zeta\left(\mathrm{3}\right)\:−\:\frac{\pi^{\mathrm{3}} }{\mathrm{1024}}\:+\:\frac{\pi^{\mathrm{3}} }{\mathrm{324}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$

Question Number 221454 Answers: 2 Comments: 0

Prove ∫^( +∞) _( 1) (x/((−1 + 3x^2 − 3x^4 + 2x^6 )(ln(x−1) − 2 ln x + ln(x+1)))) dx = − ((ln 3)/4)

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{Prove} \\ $$$$\:\:\underset{\:\mathrm{1}} {\int}^{\:+\infty} \:\frac{{x}}{\left(−\mathrm{1}\:+\:\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{3}{x}^{\mathrm{4}} \:+\:\mathrm{2}{x}^{\mathrm{6}} \right)\left(\mathrm{ln}\left({x}−\mathrm{1}\right)\:−\:\mathrm{2}\:\mathrm{ln}\:{x}\:+\:\mathrm{ln}\left({x}+\mathrm{1}\right)\right)}\:\:\mathrm{d}{x}\:=\:−\:\frac{\mathrm{ln}\:\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221453 Answers: 2 Comments: 0

Question Number 221447 Answers: 1 Comments: 0

if Π_(i=1) ^n (x + r_i ) ≡ Σ_(j=0) ^n a_j x^(n−i) show that ; Σ_(i=1) ^n tan^(−1) r_i = tan^(−1) ((a_1 + a_3 + a_5 − ∙∙∙)/(a_0 − a_2 + a_4 −∙∙∙)) and Σ_(i=1) ^n tanh^(−1) r_i = tanh^(−1) ((a_1 + a_3 + a_5 + ∙∙∙)/(a_0 + a_2 + a_4 + ∙∙∙))

$$ \\ $$$$\:\:\:\:\mathrm{if}\:\:\:\:\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\:\left({x}\:+\:{r}_{{i}} \right)\:\equiv\:\underset{{j}=\mathrm{0}} {\overset{{n}} {\sum}}\:{a}_{{j}} {x}^{{n}−{i}} \: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{show}\:\mathrm{that}\:; \\ $$$$\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tan}^{−\mathrm{1}} \:{r}_{{i}} \:=\:\mathrm{tan}^{−\mathrm{1}} \:\frac{{a}_{\mathrm{1}} +\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{5}} \:−\:\centerdot\centerdot\centerdot}{{a}_{\mathrm{0}} \:−\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{4}} \:−\centerdot\centerdot\centerdot}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{and} \\ $$$$\:\:\:\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\:\mathrm{tanh}^{−\mathrm{1}} \:{r}_{{i}} \:=\:\mathrm{tanh}^{−\mathrm{1}} \:\frac{{a}_{\mathrm{1}} \:+\:{a}_{\mathrm{3}} \:+\:{a}_{\mathrm{5}} \:+\:\centerdot\centerdot\centerdot}{{a}_{\mathrm{0}} \:+\:{a}_{\mathrm{2}} \:+\:{a}_{\mathrm{4}} \:+\:\centerdot\centerdot\centerdot}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221444 Answers: 0 Comments: 3

Question Number 221438 Answers: 0 Comments: 0

∫_( 1) ^∞ ((ln(ln x))/(1−2x cos θ + x^2 )) dx ; for all θ ∈ (−π , π)

$$ \\ $$$$\:\:\:\int_{\:\mathrm{1}} ^{\infty} \frac{\mathrm{ln}\left(\mathrm{ln}\:{x}\right)}{\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta\:+\:{x}^{\mathrm{2}} }\:{dx}\:;\:\mathrm{for}\:\mathrm{all}\:\theta\:\in\:\left(−\pi\:,\:\pi\right)\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221430 Answers: 1 Comments: 0

Question Number 221417 Answers: 0 Comments: 0

Let S be delimited by the equations x=0; y=0 ; z=0 and x+y+z=0 Find the flux of vector field V(x,y,z)=(x,y,x^2 +y^2 ) through S

$${Let}\:\:{S}\:{be}\:{delimited}\:{by}\:{the}\:{equations}\: \\ $$$${x}=\mathrm{0};\:{y}=\mathrm{0}\:;\:{z}=\mathrm{0}\:{and}\:{x}+{y}+{z}=\mathrm{0} \\ $$$${Find}\:{the}\:{flux}\:{of}\:{vector}\:{field}\: \\ $$$${V}\left({x},{y},{z}\right)=\left({x},{y},{x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\:{through}\:{S} \\ $$

Question Number 221416 Answers: 0 Comments: 0

ex3. prove f^((n)) (α)=((n!)/(2πi)) ∮_( ∂S) ((f(z))/((z−α)^(n+1) )) dz ex4. Let z_0 be any point interior to a positively oriented simple closed contour C show that a. ∮_C (dz/(z−z_0 ))=2πi b. ∮_( C) (dz/((z−z_0 )^(n+1) ))=0 , n∈R^+ ex 5. Let C be any simple closed contour, described in the positive sense in the z plane and write g(z)= ∮_( C) ((s^3 +2s)/((s−z)^3 )) ds show that g(z)=6πi when z is inside C and that g(z)=0 when z is outside

$$\mathrm{ex3}. \\ $$$$\mathrm{prove} \\ $$$${f}^{\left({n}\right)} \left(\alpha\right)=\frac{{n}!}{\mathrm{2}\pi\boldsymbol{{i}}}\:\oint_{\:\partial{S}} \:\frac{{f}\left({z}\right)}{\left({z}−\alpha\right)^{{n}+\mathrm{1}} }\:\mathrm{d}{z} \\ $$$$\mathrm{ex4}. \\ $$$$\mathrm{Let}\:{z}_{\mathrm{0}} \:\mathrm{be}\:\mathrm{any}\:\mathrm{point}\:\mathrm{interior}\:\mathrm{to}\:\mathrm{a}\:\mathrm{positively} \\ $$$$\mathrm{oriented}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}\:\mathcal{C} \\ $$$$\mathrm{show}\:\mathrm{that} \\ $$$${a}.\:\oint_{{C}} \:\frac{\mathrm{d}{z}}{{z}−{z}_{\mathrm{0}} }=\mathrm{2}\pi\boldsymbol{{i}} \\ $$$$\mathrm{b}.\:\oint_{\:{C}} \frac{\mathrm{d}{z}}{\left({z}−{z}_{\mathrm{0}} \right)^{{n}+\mathrm{1}} }=\mathrm{0}\:,\:{n}\in\mathbb{R}^{+} \\ $$$$\mathrm{ex}\:\mathrm{5}. \\ $$$$\mathrm{Let}\:\mathcal{C}\:\mathrm{be}\:\mathrm{any}\:\mathrm{simple}\:\mathrm{closed}\:\mathrm{contour}, \\ $$$$\mathrm{described}\:\mathrm{in}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{sense}\:\mathrm{in}\:\mathrm{the}\:{z}\:\mathrm{plane} \\ $$$$\mathrm{and}\:\mathrm{write}\: \\ $$$$\mathrm{g}\left({z}\right)=\:\oint_{\:\mathcal{C}} \:\frac{{s}^{\mathrm{3}} +\mathrm{2}{s}}{\left({s}−{z}\right)^{\mathrm{3}} }\:\mathrm{d}{s} \\ $$$$\mathrm{show}\:\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{6}\pi\boldsymbol{{i}}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{inside}\:\mathcal{C}\:\mathrm{and} \\ $$$$\mathrm{that}\:\mathrm{g}\left({z}\right)=\mathrm{0}\:\mathrm{when}\:{z}\:\mathrm{is}\:\mathrm{outside} \\ $$

Question Number 221415 Answers: 1 Comments: 0

∫ dz [−(1/π)((2/z))^ν ∙Σ_(k=0) ^(ν−1) ((Γ(ν−k))/(k!))((z/2))^(2k) +(2/π)ln((1/2)z)J_ν (z)−(1/π)((z/2))^ν Σ_(k=0) ^∞ (((−1)^k (ψ^((0)) (k+ν+1)+ψ^((0)) (k+1)))/(k!(k+ν)!))((z/2))^(2k) ]

$$\int\:\:\mathrm{d}{z}\:\left[−\frac{\mathrm{1}}{\pi}\left(\frac{\mathrm{2}}{{z}}\right)^{\nu} \centerdot\underset{{k}=\mathrm{0}} {\overset{\nu−\mathrm{1}} {\sum}}\:\frac{\Gamma\left(\nu−{k}\right)}{{k}!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} +\frac{\mathrm{2}}{\pi}\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}{z}\right){J}_{\nu} \left({z}\right)−\frac{\mathrm{1}}{\pi}\left(\frac{{z}}{\mathrm{2}}\right)^{\nu} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{k}} \left(\psi^{\left(\mathrm{0}\right)} \left({k}+\nu+\mathrm{1}\right)+\psi^{\left(\mathrm{0}\right)} \left({k}+\mathrm{1}\right)\right)}{{k}!\left({k}+\nu\right)!}\left(\frac{{z}}{\mathrm{2}}\right)^{\mathrm{2}{k}} \right] \\ $$

Question Number 221413 Answers: 7 Comments: 0

Question Number 221411 Answers: 1 Comments: 0

Question Number 221407 Answers: 0 Comments: 0

A and B are two angles such that 0^0 <B<A<90^0 then prove geometrycaly that cos (A+B)=cos Acos B−sin Asin B

$${A}\:{and}\:{B}\:{are}\:{two}\:{angles}\:{such}\:{that}\:\mathrm{0}^{\mathrm{0}} <{B}<{A}<\mathrm{90}^{\mathrm{0}} \:{then}\:{prove}\:{geometrycaly}\:{that}\: \\ $$$$\mathrm{cos}\:\left({A}+{B}\right)=\mathrm{cos}\:{A}\mathrm{cos}\:{B}−\mathrm{sin}\:{A}\mathrm{sin}\:{B} \\ $$

Question Number 221406 Answers: 1 Comments: 1

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Question Number 221400 Answers: 1 Comments: 0

∫∫∫_(0≤x≤y≤z≤1) [(y − x)^2 (z − y)^2 (z − x)^2 ] dxdydz

$$ \\ $$$$\:\:\:\:\:\:\:\int\int\int_{\mathrm{0}\leqslant{x}\leqslant{y}\leqslant{z}\leqslant\mathrm{1}} \:\left[\left({y}\:−\:{x}\right)^{\mathrm{2}} \left({z}\:−\:{y}\right)^{\mathrm{2}} \left({z}\:−\:{x}\right)^{\mathrm{2}} \right]\:{dxdydz}\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 221399 Answers: 0 Comments: 0

let a,b ≥ 0 and a + b + ab = 3 show that; ((38)/(55)) ≤ (1/(a^2 + 2)) + (1/(b^2 +2)) + (1/(a^2 + b^2 + 1)) ≤ 1 , ((463)/(812)) ≤ (1/(a^3 + 2)) + (1/(b^3 + 2)) + (1/(a^3 + b^3 + 1)) ≤ 1 , ((193)/(308)) ≤ (1/(a^2 + 2)) + (1/(b^3 + 2)) + (1/(a^3 + b^2 + 1)) ≤1 , and ((463)/(812)) ≤ (1/(a^2 + 2)) + (1/(b^3 + 2)) + (1/(a^2 + b^3 + 1)) ≤ ((11)/(10))

$$ \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{let}}\:\boldsymbol{{a}},\boldsymbol{{b}}\:\geqslant\:\mathrm{0}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{{a}}\:+\:\boldsymbol{{b}}\:+\:\boldsymbol{{ab}}\:=\:\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{show}}\:\boldsymbol{\mathrm{that}}; \\ $$$$\:\:\frac{\mathrm{38}}{\mathrm{55}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{2}} \:+\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\mathrm{1}}\:\leqslant\:\mathrm{1}\:\:,\:\:\:\:\:\:\: \\ $$$$\:\:\frac{\mathrm{463}}{\mathrm{812}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{1}}\:\leqslant\:\mathrm{1}\:\:\:,\:\:\:\:\:\:\:\: \\ $$$$\:\:\frac{\mathrm{193}}{\mathrm{308}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{3}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\mathrm{1}}\:\:\:\leqslant\mathrm{1}\:\:,\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{and}} \\ $$$$\:\:\frac{\mathrm{463}}{\mathrm{812}}\:\leqslant\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{2}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{3}} \:+\:\mathrm{1}}\:\leqslant\:\frac{\mathrm{11}}{\mathrm{10}}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\: \\ $$

Question Number 221397 Answers: 0 Comments: 1

Question Number 221393 Answers: 0 Comments: 0

a, b, c are complex number and ∣a∣ = ∣b∣=∣c∣= 1 and (a^2 /(bc))+(b^2 /(ac)) +(c^2 /(ab)) = −1 where ∣.∣ is modules function then ∣a+b+c∣ can be (A) 0 (B) 1 (C) (3/2) (D) 2

$$\:\:\:\:\:{a},\:{b},\:{c}\:{are}\:{complex}\:{number}\:{and}\: \\ $$$$\:\:\:\mid{a}\mid\:=\:\mid{b}\mid=\mid{c}\mid=\:\mathrm{1}\:{and}\:\:\frac{{a}^{\mathrm{2}} }{{bc}}+\frac{{b}^{\mathrm{2}} }{{ac}}\:+\frac{{c}^{\mathrm{2}} }{{ab}}\:=\:−\mathrm{1} \\ $$$$\:\:\:\:{where}\:\mid.\mid\:{is}\:\:{modules}\:{function} \\ $$$${then}\:\mid{a}+{b}+{c}\mid\:{can}\:{be}\: \\ $$$$\left({A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\left({B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\left({C}\right)\:\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\left({D}\right)\:\mathrm{2} \\ $$

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