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Question Number 219657    Answers: 1   Comments: 0

prove; Π_(n=1) ^∞ (((5n−2)(5n−3))/((5n−1)(5n−4))) = ϕ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}\:=\:\varphi \\ $$$$ \\ $$

Question Number 219651    Answers: 1   Comments: 0

Solve x^2 y^((2)) (x)+xy^((1)) (x)+(x^2 −ν^2 )y(x)=e^(−kx)

$$\mathrm{Solve} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)={e}^{−{kx}} \\ $$

Question Number 219649    Answers: 0   Comments: 0

Solve y^((2)) (t)−t∙y(t)=0

$$\mathrm{Solve}\:{y}^{\left(\mathrm{2}\right)} \left({t}\right)−{t}\centerdot{y}\left({t}\right)=\mathrm{0} \\ $$

Question Number 219642    Answers: 1   Comments: 0

Question Number 219685    Answers: 2   Comments: 0

Question Number 219637    Answers: 0   Comments: 0

∫_0 ^( ∞) ((J_ν (s)e^(−μs) )/( (√(s^2 +R^2 ))))ds , (ν,μ∈R^+ , R∈R^+ \{0})

$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{J}_{\nu} \left({s}\right){e}^{−\mu{s}} }{\:\sqrt{{s}^{\mathrm{2}} +{R}^{\mathrm{2}} }}\mathrm{d}{s}\:,\:\left(\nu,\mu\in\mathbb{R}^{+} \:,\:\mathrm{R}\in\mathbb{R}^{+} \backslash\left\{\mathrm{0}\right\}\right) \\ $$

Question Number 219634    Answers: 1   Comments: 0

Question Number 219624    Answers: 1   Comments: 0

Question Number 219625    Answers: 0   Comments: 0

Determine all real numbers x that statisfy the following inequality; (√(x^4 +1))+(√(x^4 +4x^2 +4)) ≤ (√((x^2 +1)(x^2 +4))) + ∣x∣(√(x^2 +3))

$$ \\ $$$$\:\mathrm{Determine}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:{x}\: \\ $$$$\:\:\:\mathrm{that}\:\mathrm{statisfy}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}; \\ $$$$\:\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}\:\leqslant\:\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:+\:\mid{x}\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 219621    Answers: 2   Comments: 1

Question Number 219620    Answers: 1   Comments: 0

Prove; ∫_( 0) ^( ∞) ((50x^8 )/(x^(20) +2x^(10) +1)) dx = φπ

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{Prove};\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{50}{x}^{\mathrm{8}} }{{x}^{\mathrm{20}} +\mathrm{2}{x}^{\mathrm{10}} +\mathrm{1}}\:{dx}\:=\:\phi\pi \\ $$$$ \\ $$

Question Number 219619    Answers: 2   Comments: 0

∫_0 ^∞ (x^2 +1)^(−1/2) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$$ \\ $$

Question Number 219618    Answers: 0   Comments: 0

Prove; ∫_0 ^( 1) ((ln ln (1/x))/((1+x)^2 )) dx = (1/2)(ln((π/2))−γ)

$$ \\ $$$$\:\:\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{{ln}\:{ln}\:\frac{\mathrm{1}}{{x}}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{\pi}{\mathrm{2}}\right)−\gamma\right) \\ $$$$ \\ $$

Question Number 219617    Answers: 0   Comments: 0

ok, let′s all answer questions from anywhere on the www using the given results from the sources or wolframalpha or any AI available. this promises great fun!

$$\mathrm{ok},\:\mathrm{let}'\mathrm{s}\:\mathrm{all}\:\mathrm{answer}\:\mathrm{questions}\:\mathrm{from}\:\mathrm{anywhere} \\ $$$$\mathrm{on}\:\mathrm{the}\:{www}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given}\:\mathrm{results}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{sources}\:\mathrm{or}\:{wolframalpha}\:\mathrm{or}\:\mathrm{any}\:\mathrm{AI} \\ $$$$\mathrm{available}.\:\mathrm{this}\:\mathrm{promises}\:\mathrm{great}\:\mathrm{fun}! \\ $$

Question Number 219606    Answers: 2   Comments: 0

prove that for positive real numbers a,b,c, the following inequality holds; (a^2 /(b + c)) + (b^2 /(c + a)) + (c^2 /(a + b)) ≥ ((a + b + c)/2)

$$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 219602    Answers: 1   Comments: 2

∫_0 ^( ∞) dt e^(−ikt) J_(−(2/3)) (t)−∫_0 ^( ∞) dt e^(−ikt) Y_(−(2/3)) (t)=??

$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {J}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)−\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {Y}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)=?? \\ $$

Question Number 222520    Answers: 0   Comments: 3

Question Number 222519    Answers: 0   Comments: 0

Prove: ∫_0 ^(+∞) (√(cosh x))−(√(sinh x))dx=((2−(√2))/( (√π)))Γ^2 ((3/4))

$$\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{+\infty} \sqrt{\mathrm{cosh}\:{x}}−\sqrt{\mathrm{sinh}\:{x}}{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\pi}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$

Question Number 219597    Answers: 1   Comments: 0

Evaluate integral by Complex integral method ∫_0 ^( 2π) (1/(a+b∙cos(nθ))) dθ

$$\mathrm{Evaluate}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{{a}+{b}\centerdot\mathrm{cos}\left({n}\theta\right)}\:\mathrm{d}\theta \\ $$

Question Number 219591    Answers: 0   Comments: 0

LT{((Ai^((1)) (−((3/2))^(2/3) z^(2/3) )+(√3)Bi^((1)) (−((3/2))^(2/3) z^(2/3) )/(^3 (√2)∙^6 (√3)z^(2/3) ))}=??? LT{∗}=∫_0 ^( ∞) e^(−zt) ∗ Ai(x) and Bi(x) Airy Function f^((1)) (z) is ((d )/dz)f(z)

$$\boldsymbol{\mathrm{LT}}\left\{\frac{\mathrm{Ai}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right)+\sqrt{\mathrm{3}}\mathrm{Bi}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right.}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\centerdot^{\mathrm{6}} \sqrt{\mathrm{3}}{z}^{\mathrm{2}/\mathrm{3}} }\right\}=??? \\ $$$$\boldsymbol{\mathrm{LT}}\left\{\ast\right\}=\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{zt}} \ast \\ $$$$\mathrm{Ai}\left({x}\right)\:\mathrm{and}\:\mathrm{Bi}\left({x}\right)\:\mathrm{Airy}\:\mathrm{Function} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({z}\right)\:\mathrm{is}\:\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}{f}\left({z}\right) \\ $$

Question Number 219589    Answers: 0   Comments: 2

Evaluate; L(tan^(−1) (t−(1/t))) solution; ⇒F(s)= L(tan^(−1) (t−(1/t))) ⇔ sF(s)+(π/2)=L(((t^2 +1)/(t^4 −t^2 +1)))(s) ⇒ sF(s)+(π/2)=L(((1/2)/(t^2 −(√3) t +1))+((1/2)/(t^2 −(√3) t+1)))(s) ⇒ sF(s)+(π/2)=(1/2)L((1/((t−((√3)/2))^2 +(1/4))))(s)+(1/2)L((1/((t+((√3)/2))^2 +(1/4))))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s)+(1/2)e^(((√3)/2) s) L((1/(t^2 +((1/2))^2 )))(s) ⇒ sF(s)+(π/2)=(1/2)e^(−((√3)/2) s) (1/(1/2)) sin ((1/2) s)∗(1/s)+(1/2)e^(((√3)/2) s) (1/(1/2))sin((1/2) s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) L(sin((1/2) t))(s)∗(1/s)+e^(((√3)/2) s) L(sin((1/2) t))(s)∗(1/s) ⇒ sF(s)+(π/2)=e^(−((√3)/2) s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/(2 ))s) ((1/2)/(s^2 +1/4))∗(1/s) Final Answer; F(s)=(1/s)(e^(−((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s)+e^(((√3)/2)s) ((1/2)/(s^2 +1/4))∗(1/s))−(π/(2s))

$${Evaluate};\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{solution}; \\ $$$$\:\Rightarrow{F}\left({s}\right)=\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\Leftrightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\:\:\:\:\: \\ $$$$\mathrm{Final}\:\mathrm{Answer}; \\ $$$$\:\:\:{F}\left({s}\right)=\frac{\mathrm{1}}{{s}}\left({e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\right)−\frac{\pi}{\mathrm{2}{s}}\:\: \\ $$$$ \\ $$

Question Number 219586    Answers: 1   Comments: 0

Integrate : ∫(x^2 /( (√(x^2 − x)) + 1))dx.

$${Integrate}\:: \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:−\:{x}}\:+\:\mathrm{1}}{dx}. \\ $$

Question Number 219587    Answers: 1   Comments: 1

Question Number 219600    Answers: 1   Comments: 0

∫_( 1) ^( 2) [x]^x dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \left[{x}\right]^{{x}} \:{dx} \\ $$$$ \\ $$

Question Number 219581    Answers: 0   Comments: 0

Question Number 219580    Answers: 0   Comments: 0

∫_0 ^( ∞) J_ν (kt)e^(−t) dt

$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left({kt}\right){e}^{−{t}} \:\mathrm{d}{t} \\ $$

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