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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

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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