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Question Number 196408 by Erico last updated on 24/Aug/23
Calcul ∫^( +∞) _( 0) ((lnt)/( (√t)(1+t^2 )))dt
$$\mathrm{Calcul}\:\underset{\:\mathrm{0}} {\int}^{\:+\infty} \frac{\mathrm{lnt}}{\:\sqrt{\mathrm{t}}\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)}\mathrm{dt} \\ $$
Answered by qaz last updated on 24/Aug/23
∫_0 ^∞ ((lnt)/( (√t)(1+t^2 )))dt=(∂/∂a)∣_(a=−1/2) ∫_0 ^∞ (t^a /(1+t^2 ))dt =−(∂/∂a)∣_(a=−1/2) (π/2)csc(((a+1)/2)π)=(π^2 /4)cot (((a+1)/2)π)csc(((a+1)/2)π)∣_(a=−1/2) =(((√2)π^2 )/4)
$$\int_{\mathrm{0}} ^{\infty} \frac{{lnt}}{\:\sqrt{{t}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}=\frac{\partial}{\partial{a}}\mid_{{a}=−\mathrm{1}/\mathrm{2}} \int_{\mathrm{0}} ^{\infty} \frac{{t}^{{a}} }{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=−\frac{\partial}{\partial{a}}\mid_{{a}=−\mathrm{1}/\mathrm{2}} \frac{\pi}{\mathrm{2}}{csc}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\pi\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\mathrm{cot}\:\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\pi\right){csc}\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\pi\right)\mid_{{a}=−\mathrm{1}/\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{2}}\pi^{\mathrm{2}} }{\mathrm{4}} \\ $$
Answered by Mathspace last updated on 25/Aug/23
I=∫_0 ^∞ ((lnt)/( (√t)(1+t^2 )))dt ((√t)=u) =∫_0 ^∞ ((2lnu)/(u(1+u^4 )))(2u)du =4∫_0 ^∞ ((lnu)/(1+u^4 ))du (u^4 =x) =4∫_0 ^∞ ((ln(x^(1/4) ))/(1+x))×(1/4)x^((1/4)−1) dx =(1/4)∫_0 ^∞ ((x^((1/4)−1) lnx)/(1+x))dx f(a)=∫_0 ^∞ (x^(a−1) /(1+x))dx −1<a<1 f^′ (a)=∫_0 ^∞ (x^(a−1) /(1+x))lnx dx ⇒ 4I=f^′ ((1/4)) but f(a)=(π/(sin(πa))) ⇒f^′ (a)=π.((−πcos(πa))/(sin^2 (πa))) =−π^2 ((cos(πa))/(sin^2 (πa))) ⇒ f^′ ((1/4))=−π^2 ((cos((π/4)))/(sin^2 ((π/4)))) =−π^2 .(1/( (√2).((1/2))))=−π^2 (√2) I=(1/4)f^′ ((1/4))⇒I=−((√2)/4)π^2
$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{lnt}}{\:\sqrt{{t}}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{dt}\:\:\:\:\left(\sqrt{{t}}={u}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{lnu}}{{u}\left(\mathrm{1}+{u}^{\mathrm{4}} \right)}\left(\mathrm{2}{u}\right){du} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{lnu}}{\mathrm{1}+{u}^{\mathrm{4}} }{du}\:\:\:\left({u}^{\mathrm{4}} ={x}\right) \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left({x}^{\frac{\mathrm{1}}{\mathrm{4}}} \right)}{\mathrm{1}+{x}}×\frac{\mathrm{1}}{\mathrm{4}}{x}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {lnx}}{\mathrm{1}+{x}}{dx} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}}{dx}\:\:\:\:\:\:\:\:\:−\mathrm{1}<{a}<\mathrm{1} \\ $$$${f}^{'} \left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{a}−\mathrm{1}} }{\mathrm{1}+{x}}{lnx}\:{dx}\:\Rightarrow \\ $$$$\mathrm{4}{I}={f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\:{but}\:{f}\left({a}\right)=\frac{\pi}{{sin}\left(\pi{a}\right)} \\ $$$$\Rightarrow{f}^{'} \left({a}\right)=\pi.\frac{−\pi{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)} \\ $$$$=−\pi^{\mathrm{2}} \frac{{cos}\left(\pi{a}\right)}{{sin}^{\mathrm{2}} \left(\pi{a}\right)}\:\Rightarrow \\ $$$${f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{2}} \frac{{cos}\left(\frac{\pi}{\mathrm{4}}\right)}{{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}\right)} \\ $$$$=−\pi^{\mathrm{2}} .\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}.\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}=−\pi^{\mathrm{2}} \sqrt{\mathrm{2}} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}}{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Rightarrow{I}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\pi^{\mathrm{2}} \\ $$

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