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Question Number 218766 Answers: 0 Comments: 0

Σ_(ℓ=1) ^∞ (1/ℓ)J_ν (ℓt)=??

$$\underset{\ell=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\ell}{J}_{\nu} \left(\ell{t}\right)=?? \\ $$

Question Number 218765 Answers: 1 Comments: 0

solve ∫_0 ^( ∞) J_ν (kt)e^(−wt) dt=g_(ν,k) (w)

$$\mathrm{solve} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \:\:{J}_{\nu} \left({kt}\right){e}^{−{wt}} \mathrm{d}{t}=\mathrm{g}_{\nu,{k}} \left({w}\right) \\ $$

Question Number 218749 Answers: 1 Comments: 0

Question Number 218748 Answers: 1 Comments: 0

∫_0 ^∞ ((x^2 cos x)/(cosh 2x−cos x))−((2x^2 )/(e^(4x) −2e^(2x) cos x+1))dx,lemma:Σ_(k=1) ^(+∞) ((cos kx)/p^k )=((p cos x−1)/(p^2 −2p cos x+1)),p>1

$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{2}} \mathrm{cos}\:{x}}{\mathrm{cosh}\:\mathrm{2}{x}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{e}^{\mathrm{4}{x}} −\mathrm{2}{e}^{\mathrm{2}{x}} \mathrm{cos}\:{x}+\mathrm{1}}{dx},\mathrm{lemma}:\underset{{k}=\mathrm{1}} {\overset{+\infty} {\sum}}\frac{\mathrm{cos}\:{kx}}{{p}^{{k}} }=\frac{\mathrm{p}\:\mathrm{cos}\:{x}−\mathrm{1}}{{p}^{\mathrm{2}} −\mathrm{2}{p}\:\mathrm{cos}\:{x}+\mathrm{1}},{p}>\mathrm{1} \\ $$

Question Number 218739 Answers: 2 Comments: 0

Question Number 218738 Answers: 2 Comments: 0

Question Number 218737 Answers: 2 Comments: 0

Question Number 218736 Answers: 5 Comments: 0

Question Number 218735 Answers: 5 Comments: 0

Question Number 218734 Answers: 7 Comments: 0

Question Number 218733 Answers: 4 Comments: 1

Question Number 218719 Answers: 0 Comments: 0

Question Number 218713 Answers: 1 Comments: 3

Question Number 218709 Answers: 2 Comments: 0

∫_0 ^(π/2) (dθ/( (√(sinθ +cosθ))))

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \:\frac{{d}\theta}{\:\sqrt{{sin}\theta\:+{cos}\theta}} \\ $$$$ \\ $$

Question Number 218703 Answers: 3 Comments: 0

Prove; lim_(x→0) ((x − sin x)/x^3 ) = (1/6)

$$ \\ $$$$\:\:\:\:{Prove};\:{lim}_{{x}\rightarrow\mathrm{0}} \:\frac{{x}\:−\:{sin}\:{x}}{{x}^{\mathrm{3}} }\:=\:\frac{\mathrm{1}}{\mathrm{6}}\:\: \\ $$$$ \\ $$

Question Number 218676 Answers: 2 Comments: 0

Question Number 218675 Answers: 4 Comments: 0

Question Number 218674 Answers: 3 Comments: 0

Question Number 218673 Answers: 3 Comments: 1

Question Number 218672 Answers: 2 Comments: 0

Question Number 218662 Answers: 4 Comments: 0

Prove: ∫^∞ _0 ((sin(x))/x) dx = (π/2)

$$\: \\ $$$$\:\:\:\:{Prove}:\:\:\:\:\underset{\mathrm{0}} {\int}^{\infty} \:\frac{{sin}\left({x}\right)}{{x}}\:{dx}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$

Question Number 218658 Answers: 2 Comments: 0

Question Number 218769 Answers: 1 Comments: 0

Fourier Series f(θ)=e^(izsin(θ))

$$\mathrm{Fourier}\:\mathrm{Series}\:{f}\left(\theta\right)={e}^{\boldsymbol{{i}}{z}\mathrm{sin}\left(\theta\right)} \\ $$

Question Number 218656 Answers: 2 Comments: 0

resolve the equation with unknow p P is polynom 1) P(X^2 )=(X^2 +1)P(X) 2) P 0P =P

$${resolve}\:{the}\:{equation}\:{with}\:{unknow}\:{p} \\ $$$${P}\:\:{is}\:{polynom}\: \\ $$$$\left.\mathrm{1}\right)\:{P}\left({X}^{\mathrm{2}} \right)=\left({X}^{\mathrm{2}} +\mathrm{1}\right){P}\left({X}\right) \\ $$$$\left.\mathrm{2}\right)\:{P}\:\mathrm{0}{P}\:={P} \\ $$

Question Number 218652 Answers: 0 Comments: 4

Angle of incidence is 40° if the direction of the incidence ray is constant and the mirror is rotated through 20° the angle between the new reflected ray and new normal is

Question Number 218651 Answers: 1 Comments: 0

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