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Question Number 42505 by maxmathsup by imad last updated on 26/Aug/18

calculate ∫_0 ^∞ ((ln(t))/((1+t)(√t))) dt .

$${calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}\:{dt}\:. \\ $$

Commented by maxmathsup by imad last updated on 29/Aug/18

let f(a) = ∫_0 ^∞ (t^(a−1) /(1+t))dt with 0<a<1 we know that f(a)=(π/(sin(πa))) we have f(a) = ∫_0 ^∞ (e^((a−1)ln(t)) /(1+t))dt ⇒f^′ (a) =∫_0 ^∞ ((ln(t) t^(a−1) )/(1+t))dt but ∫_0 ^∞ ((ln(t)t^(a−1) )/(1+t))dt =((π/(sin(πa))))′ =π((−π cos(πa))/(sin^2 (πa))) =−π^2 ((cos(πa))/(sin^2 (πa))) we have ∫_0 ^∞ ((ln(t))/((1+t)(√t)))dt =∫_0 ^∞ ((ln(t)t^(1−(1/2)) )/(1+t))dt =f^′ ((1/2)) =−π^2 ((cos((π/2)))/(sin^2 ((π/2)))) =0 .

$${let}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:{with}\:\:\mathrm{0}<{a}<\mathrm{1}\:\:\:{we}\:{know}\:{that}\:{f}\left({a}\right)=\frac{\pi}{{sin}\left(\pi{a}\right)} \\ $$$${we}\:{have}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{\left({a}−\mathrm{1}\right){ln}\left({t}\right)} }{\mathrm{1}+{t}}{dt}\:\Rightarrow{f}^{'} \left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({t}\right)\:{t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({t}\right){t}^{{a}−\mathrm{1}} }{\mathrm{1}+{t}}{dt}\:=\left(\frac{\pi}{{sin}\left(\pi{a}\right)}\right)'\:=\pi\frac{−\pi\:{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}\:=−\pi^{\mathrm{2}} \:\:\frac{{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$${we}\:{have}\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({t}\right){t}^{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{t}}{dt}\:={f}^{'} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:=−\pi^{\mathrm{2}} \:\:\:\frac{{cos}\left(\frac{\pi}{\mathrm{2}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)}\:=\mathrm{0}\:. \\ $$

Commented by maxmathsup by imad last updated on 29/Aug/18

another way changement (√t)=x give ∫_0 ^∞ ((ln(t))/((1+t)(√t)))dt = ∫_0 ^∞ ((ln(x^2 ))/((1+x^2 )x))(2x)dx =4 ∫_0 ^∞ ((ln(x))/(1+x^2 ))dx let I =∫_0 ^∞ ((ln(x))/(1+x^2 ))dx I =_(x=(1/u)) − ∫_0 ^∞ ((−ln(u))/(1+(1/u^2 )))(−(du/u^2 )) = −∫_0 ^∞ ((ln(u))/(1+u^2 )) =−I ⇒2I=0 ⇒I=0 so ∫_0 ^∞ ((ln(t))/((1+t)(√t)))dt =0 .

$${another}\:{way}\:\:{changement}\:\sqrt{{t}}={x}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}{dt}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right){x}}\left(\mathrm{2}{x}\right){dx}\:\:=\mathrm{4}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$${I}\:=_{{x}=\frac{\mathrm{1}}{{u}}} \:\:\:\:\:\:−\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{−{ln}\left({u}\right)}{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}\left(−\frac{{du}}{{u}^{\mathrm{2}} }\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{ln}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{2}} }\:=−{I}\:\Rightarrow\mathrm{2}{I}=\mathrm{0}\:\Rightarrow{I}=\mathrm{0}\:{so} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{ln}\left({t}\right)}{\left(\mathrm{1}+{t}\right)\sqrt{{t}}}{dt}\:=\mathrm{0}\:. \\ $$

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