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Question Number 137953 by mnjuly1970 last updated on 08/Apr/21

....Nice ... ... ... ...Calculus.... find the value of:: 𝛗=∫_1 ^( ∞) ((ln(1+x))/x^2 )dx=???

$$\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....\mathscr{N}{ice}\:...\:...\:...\:...\mathscr{C}{alculus}.... \\ $$$$\:\:\:\:\:{find}\:\:{the}\:{value}\:{of}:: \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }{dx}=??? \\ $$

Answered by Ñï= last updated on 08/Apr/21

Ο†=∫_1 ^∞ ((ln(1+x))/x^2 )dx=βˆ’βˆ«_1 ^0 ((ln(1+(1/x)))/x^2 )x^2 dx =∫_0 ^1 ln(1+x)βˆ’lnxdx ={(x+1)ln(x+1)βˆ’xlnx}_0 ^1 =2ln2

$$\phi=\int_{\mathrm{1}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}\right)}{{x}^{\mathrm{2}} }{dx}=βˆ’\int_{\mathrm{1}} ^{\mathrm{0}} \frac{{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{x}}\right)}{{x}^{\mathrm{2}} }{x}^{\mathrm{2}} {dx} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right)βˆ’{lnxdx} \\ $$$$=\left\{\left({x}+\mathrm{1}\right){ln}\left({x}+\mathrm{1}\right)βˆ’{xlnx}\right\}_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\mathrm{2}{ln}\mathrm{2} \\ $$

Commented by mnjuly1970 last updated on 08/Apr/21

very nice solution...thank you mr ...

$$\:\:{very}\:{nice}\:{solution}...{thank}\:{you} \\ $$$${mr}\:... \\ $$

Answered by mnjuly1970 last updated on 08/Apr/21

f(a)=∫_1 ^( ∞) ((ln(1+ax))/x^2 )dx f β€²(a)=∫_1 ^( ∞) (1/(x(1+ax)))dx =∫_1 ^( ∞) ((1/x) βˆ’(a/(1+ax)))dx=[ln((x/(1+ax)))]_1 ^∞ =βˆ’ln(a)+ln(1+a) ∴ f(a)=aβˆ’aln(a)+(1+a)ln(1+a)βˆ’(a)+C f(0)=0=C 𝛗=f (1)=1+2ln(2)βˆ’1=ln(4) 𝛗=ln(4)...βœ“

$$\:\:\:\:{f}\left({a}\right)=\int_{\mathrm{1}} ^{\:\infty} \frac{{ln}\left(\mathrm{1}+{ax}\right)}{{x}^{\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:{f}\:'\left({a}\right)=\int_{\mathrm{1}} ^{\:\infty} \frac{\mathrm{1}}{{x}\left(\mathrm{1}+{ax}\right)}{dx} \\ $$$$\:\:\:\:\:\:=\int_{\mathrm{1}} ^{\:\infty} \left(\frac{\mathrm{1}}{{x}}\:βˆ’\frac{{a}}{\mathrm{1}+{ax}}\right){dx}=\left[{ln}\left(\frac{{x}}{\mathrm{1}+{ax}}\right)\right]_{\mathrm{1}} ^{\infty} \\ $$$$=βˆ’{ln}\left({a}\right)+{ln}\left(\mathrm{1}+{a}\right) \\ $$$$\:\therefore\:{f}\left({a}\right)={a}βˆ’{aln}\left({a}\right)+\left(\mathrm{1}+{a}\right){ln}\left(\mathrm{1}+{a}\right)βˆ’\left({a}\right)+{C} \\ $$$$\:\:\:{f}\left(\mathrm{0}\right)=\mathrm{0}={C} \\ $$$$\:\:\:\boldsymbol{\phi}={f}\:\left(\mathrm{1}\right)=\mathrm{1}+\mathrm{2}{ln}\left(\mathrm{2}\right)βˆ’\mathrm{1}={ln}\left(\mathrm{4}\right) \\ $$$$\:\:\:\:\boldsymbol{\phi}={ln}\left(\mathrm{4}\right)...\checkmark \\ $$

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