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Question Number 222224    Answers: 0   Comments: 1

Prove:∫_0 ^1 ((ln(1−x^2 ))/x)cos(ln x)dx=1−(π/2)cosh(π/2)

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

Question Number 222218    Answers: 1   Comments: 0

∫_0 ^( π) tan^(−1) (((ln sin(x))/x)) dx

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\pi} \:\mathrm{tan}^{−\mathrm{1}} \:\left(\frac{\mathrm{ln}\:\mathrm{sin}\left({x}\right)}{{x}}\right)\:{dx} \\ $$$$ \\ $$

Question Number 222217    Answers: 0   Comments: 0

∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)=π^2 (γ+2 ln 2) Sol:∫_0 ^∞ ∫_0 ^∞ ((ln x ln y)/( (√(xy))))cos(x+y)dxdy=Re((∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx)(∫_0 ^∞ y^(−(1/2)) e^(iy) ln ydy)) ∫_0 ^∞ x^a e^(ix) dx=e^(iπ(a+1)) Γ(a+1),−1<Re a<0 ∫_0 ^∞ x^a e^(ix) dx=∫_0 ^∞ x^a e^(ix) ln xdx=(∂/∂u)[e^(iπ(a+1)/2) Γ(a+1)] =e^(iπ(a+1)) Γ(a+1)(((iπ)/2)+ψ(a+1)) a=−(1/2): ∫_0 ^∞ x^(−(1/2)) e^(ix) ln xdx=e^(iπ/1) (√π)(((iπ)/2)+ψ((1/2))) c=e^(iπ/4) (√π)(((iπ)/2)−γ−2 ln 2) e^(iπ/4) =((√2)/2)(1+i) c=(√π)∙((√2)/2)(1+i)(−γ−2 ln 2+i(π/2))=((√(2π))/2)[(γ−ln 2−(π/2))+i(−γ−2 ln 2+(π/2))] p=−γ−2 ln 2−(π/2),q=−γ−2 ln 2+(π/2) c^2 =(((√(2π))/2))^2 (p+ip)^2 =((2π)/4)(p^2 −q^2 +2ipq)=(π/2)(p^2 −q^2 +2ipq) p+q+2=(−γ−2 ln 2) p−q=π p^2 −q^2 =(p−q)(p+q)=(−π)∙2(−γ−2 ln 2)=2π(γ+2 ln 2) [2π(γ+2 ln 2)+2ipq]=π^2 (γ+2 ln 2)+iπpq Re(c^2 )=π^2 (γ+2 ln 2)

$$\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\mathrm{Sol}:\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{ln}\:{x}\:\mathrm{ln}\:{y}}{\:\sqrt{{xy}}}\mathrm{cos}\left({x}+{y}\right){dxdy}=\mathrm{Re}\left(\left(\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}\right)\left(\int_{\mathrm{0}} ^{\infty} {y}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{iy}} \mathrm{ln}\:{ydy}\right)\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} {dx}={e}^{{i}\pi\left({a}+\mathrm{1}\right)} \Gamma\left({a}+\mathrm{1}\right),−\mathrm{1}<\mathrm{Re}\:{a}<\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} {dx}=\int_{\mathrm{0}} ^{\infty} {x}^{{a}} {e}^{{ix}} \mathrm{ln}\:{xdx}=\frac{\partial}{\partial{u}}\left[{e}^{{i}\pi\left({a}+\mathrm{1}\right)/\mathrm{2}} \Gamma\left({a}+\mathrm{1}\right)\right] \\ $$$$={e}^{{i}\pi\left({a}+\mathrm{1}\right)} \Gamma\left({a}+\mathrm{1}\right)\left(\frac{{i}\pi}{\mathrm{2}}+\psi\left({a}+\mathrm{1}\right)\right) \\ $$$${a}=−\frac{\mathrm{1}}{\mathrm{2}}: \\ $$$$\int_{\mathrm{0}} ^{\infty} {x}^{−\frac{\mathrm{1}}{\mathrm{2}}} {e}^{{ix}} \mathrm{ln}\:{xdx}={e}^{{i}\pi/\mathrm{1}} \sqrt{\pi}\left(\frac{{i}\pi}{\mathrm{2}}+\psi\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\right) \\ $$$${c}={e}^{{i}\pi/\mathrm{4}} \sqrt{\pi}\left(\frac{{i}\pi}{\mathrm{2}}−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$${e}^{{i}\pi/\mathrm{4}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right) \\ $$$${c}=\sqrt{\pi}\centerdot\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}+{i}\right)\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+{i}\frac{\pi}{\mathrm{2}}\right)=\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\left[\left(\gamma−\mathrm{ln}\:\mathrm{2}−\frac{\pi}{\mathrm{2}}\right)+{i}\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\frac{\pi}{\mathrm{2}}\right)\right] \\ $$$${p}=−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}−\frac{\pi}{\mathrm{2}},{q}=−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\frac{\pi}{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{2}\pi}}{\mathrm{2}}\right)^{\mathrm{2}} \left({p}+{ip}\right)^{\mathrm{2}} =\frac{\mathrm{2}\pi}{\mathrm{4}}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2}{ipq}\right)=\frac{\pi}{\mathrm{2}}\left({p}^{\mathrm{2}} −{q}^{\mathrm{2}} +\mathrm{2}{ipq}\right) \\ $$$${p}+{q}+\mathrm{2}=\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$${p}−{q}=\pi \\ $$$${p}^{\mathrm{2}} −{q}^{\mathrm{2}} =\left({p}−{q}\right)\left({p}+{q}\right)=\left(−\pi\right)\centerdot\mathrm{2}\left(−\gamma−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)=\mathrm{2}\pi\left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$$$\left[\mathrm{2}\pi\left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)+\mathrm{2}{ipq}\right]=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right)+{i}\pi{pq} \\ $$$$\mathrm{Re}\left({c}^{\mathrm{2}} \right)=\pi^{\mathrm{2}} \left(\gamma+\mathrm{2}\:\mathrm{ln}\:\mathrm{2}\right) \\ $$

Question Number 222195    Answers: 1   Comments: 0

f(x)=x^3 +(√x)+sin x f=?

$${f}\left({x}\right)={x}^{\mathrm{3}} +\sqrt{{x}}+\mathrm{sin}\:{x} \\ $$$${f}=? \\ $$

Question Number 222193    Answers: 1   Comments: 0

Prove: (1/(sin(((3π)/8))))=2(cos((π/8))−sin((π/8))) (2/(sin(((8π)/9))))−(1/(sin(((5π)/9))))=2(√3)+4sin((π/9)) (2/(sin(((6π)/7))))+(1/(sin(((4π)/7))))=4(sin((π/7))+cos((π/(14))))

$$\mathrm{Prove}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{8}}\right)}=\mathrm{2}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{8}}\right)−\mathrm{sin}\left(\frac{\pi}{\mathrm{8}}\right)\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{8}\pi}{\mathrm{9}}\right)}−\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{5}\pi}{\mathrm{9}}\right)}=\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4sin}\left(\frac{\pi}{\mathrm{9}}\right) \\ $$$$\frac{\mathrm{2}}{\mathrm{sin}\left(\frac{\mathrm{6}\pi}{\mathrm{7}}\right)}+\frac{\mathrm{1}}{\mathrm{sin}\left(\frac{\mathrm{4}\pi}{\mathrm{7}}\right)}=\mathrm{4}\left(\mathrm{sin}\left(\frac{\pi}{\mathrm{7}}\right)+\mathrm{cos}\left(\frac{\pi}{\mathrm{14}}\right)\right) \\ $$

Question Number 222192    Answers: 1   Comments: 0

there are 32 students in a class. for each competition in a sport event in the school each class can send a team with three students. if no two students may be in the same team for more than one time, in how many different competitions can this class participate?

$${there}\:{are}\:\mathrm{32}\:{students}\:{in}\:{a}\:{class}.\:{for} \\ $$$${each}\:{competition}\:{in}\:{a}\:{sport}\:{event}\: \\ $$$${in}\:{the}\:{school}\:{each}\:{class}\:{can}\:{send} \\ $$$${a}\:{team}\:{with}\:{three}\:{students}.\:{if}\:{no} \\ $$$${two}\:{students}\:{may}\:{be}\:{in}\:{the}\:{same} \\ $$$${team}\:{for}\:{more}\:{than}\:{one}\:{time},\:{in} \\ $$$${how}\:{many}\:{different}\:{competitions}\: \\ $$$${can}\:{this}\:{class}\:{participate}? \\ $$

Question Number 222191    Answers: 0   Comments: 0

((d )/dt) ∫_( V^( 3) ) ρ_q (r,t)dV=−∮_( ∂V) J_q (r,t)∙da+∫_( V^( 3) ) S_q (r,t)dV ∫_( V^( 3) ) ((∂ρ_q (r,t))/∂t) dV=−∫_V^( 3) ▽^→ ∙J_q (r,t)dV+∫_( V^( 3) ) S_q (r,t)dV ∫_( V^( 3) ) [ (∂ρ/∂t)+▽^→ ∙J^ (r,t)−S(r,t)]dV=0 ∴((∂ρ_q (r,t))/∂t)+▽^→ ∙J_q (r,t)=S_q (r,t)

$$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\:\int_{\:{V}^{\:\mathrm{3}} } \rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V}=−\oint_{\:\partial{V}} \:\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\centerdot\mathrm{d}\boldsymbol{\mathrm{a}}+\int_{\:{V}^{\:\mathrm{3}} } \:{S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V} \\ $$$$\int_{\:{V}^{\:\mathrm{3}} } \:\frac{\partial\rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)}{\partial{t}}\:\mathrm{dV}=−\int_{{V}^{\:\mathrm{3}} } \overset{\rightarrow} {\bigtriangledown}\centerdot\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V}+\int_{\:{V}^{\:\mathrm{3}} } {S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)\mathrm{d}{V} \\ $$$$\int_{\:\mathcal{V}^{\:\mathrm{3}} } \left[\:\frac{\partial\rho}{\partial{t}}+\overset{\rightarrow} {\bigtriangledown}\centerdot\boldsymbol{\mathrm{J}}^{\:} \left(\boldsymbol{\mathrm{r}},{t}\right)−{S}\left(\boldsymbol{\mathrm{r}},{t}\right)\right]\mathrm{d}{V}=\mathrm{0} \\ $$$$\therefore\frac{\partial\rho_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)}{\partial{t}}+\overset{\rightarrow} {\bigtriangledown}\centerdot\boldsymbol{\mathrm{J}}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right)={S}_{{q}} \left(\boldsymbol{\mathrm{r}},{t}\right) \\ $$

Question Number 222184    Answers: 2   Comments: 0

f(2) = 8 f(3) = 5 ∫_2 ^( 3) (f(x) + f^′ (x)∙x) dx = ?

$$\mathrm{f}\left(\mathrm{2}\right)\:=\:\mathrm{8} \\ $$$$\mathrm{f}\left(\mathrm{3}\right)\:=\:\mathrm{5} \\ $$$$\int_{\mathrm{2}} ^{\:\mathrm{3}} \:\left(\mathrm{f}\left(\mathrm{x}\right)\:+\:\mathrm{f}\:^{'} \left(\mathrm{x}\right)\centerdot\mathrm{x}\right)\:\mathrm{dx}\:=\:? \\ $$

Question Number 222176    Answers: 1   Comments: 0

Question Number 222175    Answers: 1   Comments: 0

solve (e^(2y) −y)cosx(dy/dx)=e^y sin2x klipto−quanta

$$\boldsymbol{\mathrm{solve}} \\ $$$$\left(\boldsymbol{\mathrm{e}}^{\mathrm{2}\boldsymbol{\mathrm{y}}} −\boldsymbol{\mathrm{y}}\right)\boldsymbol{\mathrm{cosx}}\frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}=\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{y}}} \boldsymbol{\mathrm{sin}}\mathrm{2}\boldsymbol{\mathrm{x}} \\ $$$$\boldsymbol{\mathrm{klipto}}−\boldsymbol{\mathrm{quanta}} \\ $$

Question Number 222172    Answers: 2   Comments: 0

(4/5) > (8/(3x − 6)) > (2/9) find: x = ?

$$\frac{\mathrm{4}}{\mathrm{5}}\:>\:\frac{\mathrm{8}}{\mathrm{3x}\:−\:\mathrm{6}}\:>\:\frac{\mathrm{2}}{\mathrm{9}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\mathrm{x}}\:=\:? \\ $$

Question Number 222151    Answers: 3   Comments: 8

Question Number 222153    Answers: 1   Comments: 1

Question Number 222141    Answers: 2   Comments: 0

x^2 +((x/(2x−1)))^2 =12

$${x}^{\mathrm{2}} +\left(\frac{{x}}{\mathrm{2}{x}−\mathrm{1}}\right)^{\mathrm{2}} =\mathrm{12} \\ $$

Question Number 222142    Answers: 2   Comments: 0

a=3(√2) ,b=(1/(5^(1/6) (√6))) and x,yεR such that 3x +2y=log _a (18)^(5/4) 2x−y=log _b ((√(1080))) then find the value of 4x+5y

$${a}=\mathrm{3}\sqrt{\mathrm{2}}\:,{b}=\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{6}}} \sqrt{\mathrm{6}}}\:{and}\:{x},{y}\epsilon\mathbb{R}\:{such}\:{that} \\ $$$$\mathrm{3}{x}\:+\mathrm{2}{y}=\mathrm{log}\:_{{a}} \left(\mathrm{18}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{2}{x}−{y}=\mathrm{log}\:_{{b}} \left(\sqrt{\mathrm{1080}}\right) \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\: \\ $$$$\mathrm{4}{x}+\mathrm{5}{y} \\ $$

Question Number 222127    Answers: 0   Comments: 0

∫_0 ^( ∞) ((ln(tan(tan^(−1) (e^((1/π) tan^(−1) u) ))) )/(u^2 + 2πu + 2π^2 )) du

$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \:\frac{\mathrm{ln}\left(\mathrm{tan}\left(\mathrm{tan}^{−\mathrm{1}} \left({e}^{\frac{\mathrm{1}}{\pi}\:\mathrm{tan}^{−\mathrm{1}} \:{u}} \right)\right)\right)\:}{{u}^{\mathrm{2}} \:+\:\mathrm{2}\pi{u}\:+\:\mathrm{2}\pi^{\mathrm{2}} }\:{du} \\ $$$$ \\ $$

Question Number 222125    Answers: 4   Comments: 2

If: log_3 (5^x + (1/5^x ) + 7) ⇒ min = ?

$$\mathrm{If}:\:\:\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\frac{\mathrm{1}}{\mathrm{5}^{\boldsymbol{\mathrm{x}}} }\:+\:\mathrm{7}\right)\:\:\:\Rightarrow\:\:\:\mathrm{min}\:=\:? \\ $$

Question Number 222123    Answers: 1   Comments: 1

Question Number 222121    Answers: 2   Comments: 0

∫ acos(((cos(ϱ))/(1+2cos(ϱ)))) dϱ

$$\int\:\mathrm{acos}\left(\frac{\mathrm{cos}\left(\varrho\right)}{\mathrm{1}+\mathrm{2cos}\left(\varrho\right)}\right)\:\mathrm{d}\varrho \\ $$

Question Number 222117    Answers: 0   Comments: 0

problem 1. for a given positive integer m, find all triples(n,x,y) of positive integers,with n relatively prime to m,which satisfy (x^2 +y^2 )^m =(xy)^n hint:utilize AM&GM,diophantine eqn KLIPTO−QUANTA−OOZY

$$\boldsymbol{\mathrm{problem}}\:\mathrm{1}.\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integer}} \\ $$$$\boldsymbol{\mathrm{m}},\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{triples}}\left(\boldsymbol{\mathrm{n}},\boldsymbol{\mathrm{x}},\boldsymbol{\mathrm{y}}\right)\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{positive}}\:\boldsymbol{\mathrm{integers}},\boldsymbol{\mathrm{with}} \\ $$$$\boldsymbol{\mathrm{n}}\:\boldsymbol{\mathrm{relatively}}\:\boldsymbol{\mathrm{prime}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{m}},\boldsymbol{\mathrm{which}}\:\boldsymbol{\mathrm{satisfy}} \\ $$$$\left(\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} \right)^{\boldsymbol{\mathrm{m}}} =\left(\boldsymbol{\mathrm{xy}}\right)^{\boldsymbol{\mathrm{n}}} \\ $$$$\boldsymbol{\mathrm{hint}}:\boldsymbol{\mathrm{utilize}}\:\boldsymbol{\mathrm{AM\&GM}},\boldsymbol{\mathrm{diophantine}}\:\boldsymbol{\mathrm{eqn}} \\ $$$$\boldsymbol{\mathrm{KLIPTO}}−\boldsymbol{\mathrm{QUANTA}}−\boldsymbol{\mathrm{OOZY}} \\ $$

Question Number 222113    Answers: 0   Comments: 2

name the following compound

$$\mathrm{name}\:\mathrm{the}\:\mathrm{following}\:\mathrm{compound} \\ $$

Question Number 222105    Answers: 1   Comments: 1

Question Number 222104    Answers: 2   Comments: 0

log _4 x − log _x^2 8 = 1 x =?

$$\:\:\:\:\mathrm{log}\:_{\mathrm{4}} \:\mathrm{x}\:−\:\mathrm{log}\:_{\mathrm{x}^{\mathrm{2}} } \:\mathrm{8}\:=\:\mathrm{1} \\ $$$$\:\:\:\:\mathrm{x}\:=?\: \\ $$

Question Number 222100    Answers: 0   Comments: 0

Question Number 222097    Answers: 0   Comments: 2

Question Number 222095    Answers: 0   Comments: 0

could I consider Y_ν (z)=cot(νπ)J_ν (z)−csc(νπ)J_(−ν) (z) as ∞−∞ form limit when ν∈Z and How can i calculate Y_ν (z)=cot(νπ)J_ν (z)−csc(νπ)J_(−ν) (z)...?? lim_(α→ν) ((cot^2 (απ)J_α ^2 (z)−csc^2 (απ)J_(−α) ^( 2) (z))/(cot(απ)J_α (z)+csc(απ)J_(−α) (z)))..... lim_(α→ν) ((((∂ )/∂α)(cot^2 (απ)J_α ^2 (z)−csc^2 (απ)J_(−α) ^2 (z)))/(((∂ )/∂α)(cot(απ)J_α ^ (z)+csc(απ)J_(−α) (z))))....??.... :(

$$\mathrm{could}\:\:\mathrm{I}\:\mathrm{consider}\:\:{Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\nu\pi\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\nu\pi\right){J}_{−\nu} \left({z}\right) \\ $$$$\mathrm{as}\:\infty−\infty\:\mathrm{form}\:\mathrm{limit}\:\mathrm{when}\:\nu\in\mathbb{Z} \\ $$$$\mathrm{and}\:\mathrm{How}\:\mathrm{can}\:\mathrm{i}\:\mathrm{calculate} \\ $$$${Y}_{\nu} \left({z}\right)=\mathrm{cot}\left(\nu\pi\right){J}_{\nu} \left({z}\right)−\mathrm{csc}\left(\nu\pi\right){J}_{−\nu} \left({z}\right)...?? \\ $$$$\underset{\alpha\rightarrow\nu} {\mathrm{lim}}\:\frac{\mathrm{cot}^{\mathrm{2}} \left(\alpha\pi\right){J}_{\alpha} ^{\mathrm{2}} \left({z}\right)−\mathrm{csc}^{\mathrm{2}} \left(\alpha\pi\right){J}_{−\alpha} ^{\:\mathrm{2}} \left({z}\right)}{\mathrm{cot}\left(\alpha\pi\right){J}_{\alpha} \left({z}\right)+\mathrm{csc}\left(\alpha\pi\right){J}_{−\alpha} \left({z}\right)}..... \\ $$$$\underset{\alpha\rightarrow\nu} {\mathrm{lim}}\frac{\frac{\partial\:\:}{\partial\alpha}\left(\mathrm{cot}^{\mathrm{2}} \left(\alpha\pi\right){J}_{\alpha} ^{\mathrm{2}} \left({z}\right)−\mathrm{csc}^{\mathrm{2}} \left(\alpha\pi\right){J}_{−\alpha} ^{\mathrm{2}} \left({z}\right)\right)}{\frac{\partial\:\:}{\partial\alpha}\left(\mathrm{cot}\left(\alpha\pi\right){J}_{\alpha} ^{\:} \left({z}\right)+\mathrm{csc}\left(\alpha\pi\right){J}_{−\alpha} \left({z}\right)\right)}....??.... \\ $$$$:\left(\right. \\ $$

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