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Question Number 219657 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}; \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\:\frac{\left(\mathrm{5}{n}−\mathrm{2}\right)\left(\mathrm{5}{n}−\mathrm{3}\right)}{\left(\mathrm{5}{n}−\mathrm{1}\right)\left(\mathrm{5}{n}−\mathrm{4}\right)}\:=\:\varphi \\ $$$$ \\ $$
Question Number 219651 Answers: 1 Comments: 0
$$\mathrm{Solve} \\ $$$${x}^{\mathrm{2}} {y}^{\left(\mathrm{2}\right)} \left({x}\right)+{xy}^{\left(\mathrm{1}\right)} \left({x}\right)+\left({x}^{\mathrm{2}} −\nu^{\mathrm{2}} \right){y}\left({x}\right)={e}^{−{kx}} \\ $$
Question Number 219649 Answers: 0 Comments: 0
$$\mathrm{Solve}\:{y}^{\left(\mathrm{2}\right)} \left({t}\right)−{t}\centerdot{y}\left({t}\right)=\mathrm{0} \\ $$
Question Number 219642 Answers: 1 Comments: 0
Question Number 219685 Answers: 2 Comments: 0
Question Number 219637 Answers: 0 Comments: 0
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{{J}_{\nu} \left({s}\right){e}^{−\mu{s}} }{\:\sqrt{{s}^{\mathrm{2}} +{R}^{\mathrm{2}} }}\mathrm{d}{s}\:,\:\left(\nu,\mu\in\mathbb{R}^{+} \:,\:\mathrm{R}\in\mathbb{R}^{+} \backslash\left\{\mathrm{0}\right\}\right) \\ $$
Question Number 219634 Answers: 1 Comments: 0
Question Number 219624 Answers: 1 Comments: 0
Question Number 219625 Answers: 0 Comments: 0
$$ \\ $$$$\:\mathrm{Determine}\:\mathrm{all}\:\mathrm{real}\:\mathrm{numbers}\:{x}\: \\ $$$$\:\:\:\mathrm{that}\:\mathrm{statisfy}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}; \\ $$$$\:\:\sqrt{{x}^{\mathrm{4}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}}\:\leqslant\:\sqrt{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:+\:\mid{x}\mid\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\:\:\:\:\:\: \\ $$$$ \\ $$
Question Number 219621 Answers: 2 Comments: 1
Question Number 219620 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{Prove};\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{50}{x}^{\mathrm{8}} }{{x}^{\mathrm{20}} +\mathrm{2}{x}^{\mathrm{10}} +\mathrm{1}}\:{dx}\:=\:\phi\pi \\ $$$$ \\ $$
Question Number 219619 Answers: 2 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\infty} \left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}/\mathrm{2}} {dx} \\ $$$$ \\ $$
Question Number 219618 Answers: 0 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:{Prove}; \\ $$$$\:\:\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\:\frac{{ln}\:{ln}\:\frac{\mathrm{1}}{{x}}}{\left(\mathrm{1}+{x}\right)^{\mathrm{2}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({ln}\left(\frac{\pi}{\mathrm{2}}\right)−\gamma\right) \\ $$$$ \\ $$
Question Number 219617 Answers: 0 Comments: 0
$$\mathrm{ok},\:\mathrm{let}'\mathrm{s}\:\mathrm{all}\:\mathrm{answer}\:\mathrm{questions}\:\mathrm{from}\:\mathrm{anywhere} \\ $$$$\mathrm{on}\:\mathrm{the}\:{www}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given}\:\mathrm{results}\:\mathrm{from} \\ $$$$\mathrm{the}\:\mathrm{sources}\:\mathrm{or}\:{wolframalpha}\:\mathrm{or}\:\mathrm{any}\:\mathrm{AI} \\ $$$$\mathrm{available}.\:\mathrm{this}\:\mathrm{promises}\:\mathrm{great}\:\mathrm{fun}! \\ $$
Question Number 219606 Answers: 2 Comments: 0
$$ \\ $$$$\:\mathrm{prove}\:\mathrm{that}\:\mathrm{for}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{numbers}\:{a},{b},{c},\:\:\: \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{inequality}\:\mathrm{holds}; \\ $$$$\:\:\frac{{a}^{\mathrm{2}} }{{b}\:+\:{c}}\:+\:\frac{{b}^{\mathrm{2}} }{{c}\:+\:{a}}\:+\:\frac{{c}^{\mathrm{2}} }{{a}\:+\:{b}}\:\:\geqslant\:\frac{{a}\:+\:{b}\:+\:{c}}{\mathrm{2}} \\ $$$$ \\ $$
Question Number 219602 Answers: 1 Comments: 2
$$\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {J}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)−\int_{\mathrm{0}} ^{\:\infty} \:\mathrm{d}{t}\:{e}^{−\boldsymbol{{i}}{kt}} {Y}_{−\frac{\mathrm{2}}{\mathrm{3}}} \left({t}\right)=?? \\ $$
Question Number 222520 Answers: 0 Comments: 3
Question Number 222519 Answers: 0 Comments: 0
$$\mathrm{Prove}: \\ $$$$\int_{\mathrm{0}} ^{+\infty} \sqrt{\mathrm{cosh}\:{x}}−\sqrt{\mathrm{sinh}\:{x}}{dx}=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\pi}}\Gamma^{\mathrm{2}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right) \\ $$
Question Number 219597 Answers: 1 Comments: 0
$$\mathrm{Evaluate}\:\mathrm{integral}\:\mathrm{by}\:\mathrm{Complex}\:\mathrm{integral}\:\mathrm{method} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \:\:\frac{\mathrm{1}}{{a}+{b}\centerdot\mathrm{cos}\left({n}\theta\right)}\:\mathrm{d}\theta \\ $$
Question Number 219591 Answers: 0 Comments: 0
$$\boldsymbol{\mathrm{LT}}\left\{\frac{\mathrm{Ai}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right)+\sqrt{\mathrm{3}}\mathrm{Bi}^{\left(\mathrm{1}\right)} \left(−\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}/\mathrm{3}} {z}^{\mathrm{2}/\mathrm{3}} \right.}{\:^{\mathrm{3}} \sqrt{\mathrm{2}}\centerdot^{\mathrm{6}} \sqrt{\mathrm{3}}{z}^{\mathrm{2}/\mathrm{3}} }\right\}=??? \\ $$$$\boldsymbol{\mathrm{LT}}\left\{\ast\right\}=\int_{\mathrm{0}} ^{\:\infty} \:{e}^{−{zt}} \ast \\ $$$$\mathrm{Ai}\left({x}\right)\:\mathrm{and}\:\mathrm{Bi}\left({x}\right)\:\mathrm{Airy}\:\mathrm{Function} \\ $$$${f}^{\left(\mathrm{1}\right)} \left({z}\right)\:\mathrm{is}\:\frac{\mathrm{d}\:\:}{\mathrm{d}{z}}{f}\left({z}\right) \\ $$
Question Number 219589 Answers: 0 Comments: 2
$${Evaluate};\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{solution}; \\ $$$$\:\Rightarrow{F}\left({s}\right)=\:\mathscr{L}\left({tan}^{−\mathrm{1}} \left({t}−\frac{\mathrm{1}}{{t}}\right)\right) \\ $$$$\Leftrightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{4}} −{t}^{\mathrm{2}} +\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\mathscr{L}\left(\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}\:+\mathrm{1}}+\frac{\frac{\mathrm{1}}{\mathrm{2}}}{{t}^{\mathrm{2}} −\sqrt{\mathrm{3}}\:{t}+\mathrm{1}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}\mathscr{L}\left(\frac{\mathrm{1}}{\left({t}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}}\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right)+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left(\frac{\mathrm{1}}{{t}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }\right)\left({s}\right) \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{sin}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}}}\mathrm{sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \mathscr{L}\left({sin}\left(\frac{\mathrm{1}}{\mathrm{2}}\:{t}\right)\right)\left({s}\right)\ast\frac{\mathrm{1}}{{s}} \\ $$$$\Rightarrow\:{sF}\left({s}\right)+\frac{\pi}{\mathrm{2}}={e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\:\:\:\:\: \\ $$$$\mathrm{Final}\:\mathrm{Answer}; \\ $$$$\:\:\:{F}\left({s}\right)=\frac{\mathrm{1}}{{s}}\left({e}^{−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}+{e}^{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{s}} \:\frac{\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{1}/\mathrm{4}}\ast\frac{\mathrm{1}}{{s}}\right)−\frac{\pi}{\mathrm{2}{s}}\:\: \\ $$$$ \\ $$
Question Number 219586 Answers: 1 Comments: 0
$${Integrate}\:: \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{{x}^{\mathrm{2}} \:−\:{x}}\:+\:\mathrm{1}}{dx}. \\ $$
Question Number 219587 Answers: 1 Comments: 1
Question Number 219600 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{\:\mathrm{1}} ^{\:\mathrm{2}} \left[{x}\right]^{{x}} \:{dx} \\ $$$$ \\ $$
Question Number 219581 Answers: 0 Comments: 0
Question Number 219580 Answers: 0 Comments: 0
$$\int_{\mathrm{0}} ^{\:\infty} \:{J}_{\nu} \left({kt}\right){e}^{−{t}} \:\mathrm{d}{t} \\ $$
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