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Question Number 138799 by TheSupreme last updated on 18/Apr/21
∫_0 ^(π/2) ln(tan(x)−α)dx=... α∈C
∫0π2ln(tan(x)−α)dx=…α∈C
Answered by mathmax by abdo last updated on 19/Apr/21
Φ=∫_0 ^(π/2) ln(tanx−α)dx ⇒Φ=_(tanx=t) ∫_0 ^∞ ln(t−α)(dt/(1+t^2 )) =∫_0 ^∞ ((ln(t−α))/(1+t^2 ))dt let α=x+iy ⇒t−α =t−x−iy=(√((t−x)^2 +y^2 ))e^(iarctan(−(y/(t−x)))) ⇒Φ=∫_0 ^∞ ((ln((√((t−x)^2 +y^2 )))−iarctan((y/(t−x))))/(t^2 +1))dt =(1/2)∫_0 ^∞ ((ln((t−x)^2 +y^2 ))/(t^2 +1))dt−i∫_0 ^∞ ((arctan((y/(t−x))))/(t^2 +1))dt we have ∫_0 ^∞ ((log((t−x)^2 +y^2 ))/(t^2 +1))dt =∫_0 ^∞ ((log(t^2 −2xt +x^2 +y^2 ))/(t^2 +1))dt 𝛟(x)=∫_0 ^∞ ((log(t^2 −2xt +x^2 +y^2 ))/(t^2 +1))dt ⇒ ϕ^′ (x)=∫_0 ^∞ ((−2t+2x)/((t^2 −2xt+x^2 +y^2 )(t^2 +1)))dt we must decompose F(t)=((−2t+2x)/((t^2 −2xt+x^2 +y^2 )(t^2 +1))) .... ∫_0 ^∞ ((arctan((y/(t−x))))/(t^2 +1))dt =∫_0 ^∞ ((+^− (π/2)−arctan(((t−x)/y)))/(t^2 +1))dt =+^(− ) (π^2 /4)−∫_0 ^∞ ((arctan(t−x))/(t^2 +1))dt +(π/2)arctany Ψ(x)=∫_0 ^∞ ((arctan(t−x))/(t^2 +1))dt ⇒Ψ^′ (x)=∫_0 ^∞ ((−1)/((1+(t−x)^2 )(t^2 +1)))dt =−∫_0 ^∞ (dt/((t^2 −2xt +x^2 +1)(t^2 +1))) rest decomposition...be continued...
Φ=∫0π2ln(tanx−α)dx⇒Φ=tanx=t∫0∞ln(t−α)dt1+t2=∫0∞ln(t−α)1+t2dtletα=x+iy⇒t−α=t−x−iy=(t−x)2+y2eiarctan(−yt−x)⇒Φ=∫0∞ln((t−x)2+y2)−iarctan(yt−x)t2+1dt=12∫0∞ln((t−x)2+y2)t2+1dt−i∫0∞arctan(yt−x)t2+1dtwehave∫0∞log((t−x)2+y2)t2+1dt=∫0∞log(t2−2xt+x2+y2)t2+1dtφ(x)=∫0∞log(t2−2xt+x2+y2)t2+1dt⇒φ′(x)=∫0∞−2t+2x(t2−2xt+x2+y2)(t2+1)dtwemustdecomposeF(t)=−2t+2x(t2−2xt+x2+y2)(t2+1)….∫0∞arctan(yt−x)t2+1dt=∫0∞+−π2−arctan(t−xy)t2+1dt=+−π24−∫0∞arctan(t−x)t2+1dt+π2arctanyΨ(x)=∫0∞arctan(t−x)t2+1dt⇒Ψ′(x)=∫0∞−1(1+(t−x)2)(t2+1)dt=−∫0∞dt(t2−2xt+x2+1)(t2+1)restdecomposition…becontinued…
Commented by mathmax by abdo last updated on 19/Apr/21
sorry ∫_0 ^∞ ((+^− (π/2)−arctan(((t−x)/y)))/(t^2 +1))dt =+^− (π^2 /4)−∫_0 ^∞ ((arctan((t/y)−(x/y)))/(t^2 +1))dt so we considere Ψ(x)=∫_0 ^∞ ((arctan((t/y)−(x/y)))/(t^2 +1))dt ⇒ Ψ^′ (y)=∫_0 ^∞ ((−1)/(y(1+((t/y)−(x/y))^2 )(t^2 +1)))dt=.....
sorry∫0∞+−π2−arctan(t−xy)t2+1dt=+−π24−∫0∞arctan(ty−xy)t2+1dtsoweconsidereΨ(x)=∫0∞arctan(ty−xy)t2+1dt⇒Ψ′(y)=∫0∞−1y(1+(ty−xy)2)(t2+1)dt=…..
Commented by mathmax by abdo last updated on 19/Apr/21
let try another way Φ=∫_0 ^∞ ((log(t−α))/(1+t^2 )) dt =Ψ(α) ⇒ Ψ^′ (α) =∫_0 ^∞ ((−1)/((t−α)(t^2 +1)))dt let decompose F(t) =(1/((t−α)(t^2 +1))) ⇒F(t)=(a/(t−α)) +((bt+c)/(t^2 +1)) a=(1/(α^2 +1)) ,lim_(t→+∞) tF(t)=0 =a+b ⇒b=−(1/(α^2 +1)) F(0)=−(1/α)=−(a/α) +c ⇒1=a−αc ⇒αc=a−1 ⇒c=((a−1)/α) =(((1/(α^2 +1))−1)/α) =−(α^2 /(α(α^2 +1)))=−(α/(α^2 +1)) ⇒F(t)=(1/((α^2 +1)(t−α))) +((−(1/(α^2 +1))t −(α/(α^2 +1)))/(t^2 +1)) ⇒∫_0 ^∞ F(t)dt =(1/(α^2 +1))∫_0 ^∞ (dt/(t−α))−(1/(2(α^2 +1)))∫_0 ^∞ ((2t)/(t^(2 ) +1))dt−(α/(α^2 +1))×(π/2) =(1/(α^2 +1))[log(((t−α)/( (√(t^(2 ) +1)))))]_0 ^∞ −((πα)/(α^2 +1)) =(1/(α^2 +1))(−log(−α))−((πα)/(2(α^2 +1))) ⇒Ψ(α)=−∫ ((log(−α))/(α^2 +1))dα −(π/4)log(1+α^2 ) +C....be continued...
lettryanotherwayΦ=∫0∞log(t−α)1+t2dt=Ψ(α)⇒Ψ′(α)=∫0∞−1(t−α)(t2+1)dtletdecomposeF(t)=1(t−α)(t2+1)⇒F(t)=at−α+bt+ct2+1a=1α2+1,limt→+∞tF(t)=0=a+b⇒b=−1α2+1F(0)=−1α=−aα+c⇒1=a−αc⇒αc=a−1⇒c=a−1α=1α2+1−1α=−α2α(α2+1)=−αα2+1⇒F(t)=1(α2+1)(t−α)+−1α2+1t−αα2+1t2+1⇒∫0∞F(t)dt=1α2+1∫0∞dtt−α−12(α2+1)∫0∞2tt2+1dt−αα2+1×π2=1α2+1[log(t−αt2+1)]0∞−παα2+1=1α2+1(−log(−α))−πα2(α2+1)⇒Ψ(α)=−∫log(−α)α2+1dα−π4log(1+α2)+C….becontinued…
Answered by phanphuoc last updated on 18/Apr/21
u=tgx−> du/1+u^2 =dx I=∫_0 ^∞ ln(u+α)/u^2 +1du I=2πiRes(f(u),i)=πln(i+α)
u=tgx−>du/1+u2=dxI=∫0∞ln(u+α)/u2+1duI=2πiRes(f(u),i)=πln(i+α)

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