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Question Number 130307 by mnjuly1970 last updated on 24/Jan/21

... nice .... cslculus... evaluate ::: ∫_0 ^( ∞) ((ln (x ))/( (√x) (1+x^2 ))) dx=?

$$\:\:\:\:\:\:\:\:\:\:\:\:...\:{nice}\:\:\:....\:\:\:\:\:{cslculus}... \\ $$$$\:\:\:{evaluate}\::::\:\int_{\mathrm{0}} ^{\:\infty} \frac{{ln}\:\left({x}\:\right)}{\:\sqrt{{x}}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\:{dx}=? \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 24/Jan/21

∫_0 ^∞ ((log(x))/((1+x^2 )(√x)))dx x=u^2 =4∫_0 ^∞ ((log(u))/(1+u^4 ))du=4(((−π^2 )/4^2 )cosec((π/4))cot((π/4)))=−(π^2 /(2(√2)))

$$\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\sqrt{{x}}}{dx}\:\:\:\:\:\:\:\:\:\:\:\:{x}={u}^{\mathrm{2}} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \frac{{log}\left({u}\right)}{\mathrm{1}+{u}^{\mathrm{4}} }{du}=\mathrm{4}\left(\frac{−\pi^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} }{cosec}\left(\frac{\pi}{\mathrm{4}}\right){cot}\left(\frac{\pi}{\mathrm{4}}\right)\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$

Commented by Dwaipayan Shikari last updated on 24/Jan/21

Q130137

$${Q}\mathrm{130137} \\ $$

Commented by mnjuly1970 last updated on 24/Jan/21

thank you so much mr dwaypayan.

$${thank}\:{you}\:{so}\:{much}\:{mr}\:{dwaypayan}. \\ $$$$\:\: \\ $$

Answered by mathmax by abdo last updated on 24/Jan/21

A=∫_0 ^∞ ((lnx)/( (√x)(1+x^2 )))dx ⇒A=_((√x)=t) ∫_0 ^∞ ((ln(t^2 ))/(t(1+t^4 )))(2t)dt =4∫_0 ^∞ ((lnt)/(t^4 +1))dt wedo the changement t=u^(1/4) ⇒ A=∫_0 ^∞ (1/4) ((lnu)/(1+u))u^((1/4)−1) du =(1/4)∫_0 ^∞ ((u^((1/4)−1) lnu)/(1+u))du let f(a)=∫_0 ^∞ (t^(a−1) /(1+t))dt =∫_0 ^∞ (e^((a−1)lnt) /(1+t))dt =(π/(sin(πa))) with 0<a<1 f^′ (a)=∫_0 ^∞ ((t^(a−1) lnt)/(1+t))dt ⇒A =(1/4)f^′ ((1/4)) we have f^′ (a)=−((π^2 cos(πa))/(sin^2 (πa))) ⇒f^′ ((1/4))=−π^2 ×(((√2)/2)/(1/2))=−π^2 (√2) ⇒ A=−(π^2 /4)(√(2 ))=−((√2)/4)π^2

$$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnx}}{\:\sqrt{\mathrm{x}}\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)}\mathrm{dx}\:\:\Rightarrow\mathrm{A}=_{\sqrt{\mathrm{x}}=\mathrm{t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{ln}\left(\mathrm{t}^{\mathrm{2}} \right)}{\mathrm{t}\left(\mathrm{1}+\mathrm{t}^{\mathrm{4}} \right)}\left(\mathrm{2t}\right)\mathrm{dt} \\ $$$$=\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{lnt}}{\mathrm{t}^{\mathrm{4}} \:+\mathrm{1}}\mathrm{dt}\:\mathrm{wedo}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{t}=\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}} \:\Rightarrow \\ $$$$\mathrm{A}=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\mathrm{lnu}}{\mathrm{1}+\mathrm{u}}\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \:\mathrm{du}\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{u}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \mathrm{lnu}}{\mathrm{1}+\mathrm{u}}\mathrm{du} \\ $$$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\mathrm{lnt}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\mathrm{with}\:\mathrm{0}<\mathrm{a}<\mathrm{1} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{lnt}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{A}\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{f}^{'} \left(\mathrm{a}\right)=−\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\:\Rightarrow\mathrm{f}^{'} \left(\frac{\mathrm{1}}{\mathrm{4}}\right)=−\pi^{\mathrm{2}} \:×\frac{\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\frac{\mathrm{1}}{\mathrm{2}}}=−\pi^{\mathrm{2}} \sqrt{\mathrm{2}}\:\:\Rightarrow \\ $$$$\mathrm{A}=−\frac{\pi^{\mathrm{2}} }{\mathrm{4}}\sqrt{\mathrm{2}\:}=−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\pi^{\mathrm{2}} \\ $$

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