Prove-0-1-2-ln-2x-1-x-2-dx-pi-2-20-

Question Number 222838 by MrGaster last updated on 09/Jul/25

Prove:∫_0 ^(1/2) ((ln(2x))/( (√(1+x^2 ))))dx=−(π^2 /(20))

$$\:\:\mathrm{Prove}:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}=−\frac{\pi^{\mathrm{2}} }{\mathrm{20}} \\ $$

Answered by MrGaster last updated on 09/Jul/25

x=sinhθ⇒dx=coshθdθ,(√(1+x^2 ))=coshθ x=0⇒θ=0,x=(1/2)⇒θ=((1/2))=ln φ,φ=((1+(√5))/2) ∫_0 ^(1/2) ((ln(2x))/( (√(1+x^2 ))))dx=∫_0 ^(ln φ) ln(2sinh θ)dθ ln(2 sinh θ)=θ+ln(1−e^(−2θ) ) ∫_0 ^(ln φ) ln(2 sinh θ)dθ=∫_0 ^(ln φ) (θ+ln(1−e^(−2θ) ))dθ =[(θ^2 /2)]_0 ^(ln φ) +∫_0 ^(ln φ) ln(1−e^(−2θ) )dθ v=e^(−2θ) ⇒dθ=−(dv/(2v)),θ=0⇒v=1,θ=ln φ⇒v=φ^(−2) =((3−(√5))/2) ∫_0 ^(ln φ) ln(1−e^(−2θ) )dθ=∫_1 ^((3−(√5))/2) ln(1−v)=(1/2)∫_((3−(√5))/2) ^1 ((ln(1−v))/v)dx ((ln(1−v))/v)=−Σ_(k=1) ^∞ ,∣v∣<1 ∫_((3−(√5))/2) ^1 ((ln(1−v))/v)dv=−Σ_(k=1) ^∞ (1/k)∫_((3−(√5))/2) ^1 v^(k−1) dv=−Σ_(k=1) ^∞ (1/k)[(v^k /k)]_((3−(√5))/2) ^1 =−Σ_(k=1) ^∞ (1/k^2 )(1−(((3−(√5))/2))^k ) =−ζ(2)+ _2 (((3−(√5))/2))−,ζ(2)=(π^2 /6) _2 (((3−(√5))/2))=(π^2 /(15))−(ln φ)^2 ∫_0 ^(ln φ) ln(1−e^(−2θ) )dθ=(1/2)(−(π^2 /6)+(π^2 /(15))−(ln φ)^2 )=−(π^2 /(12))+(π^2 /(30))−(((ln φ)^2 )/2) ∫_0 ^(ln φ) ln(2 sinh θ)dθ=(((ln φ)^2 )/2)+(−(π^2 /(12))+(π^2 /(30))−(((ln φ)^2 )/2))=−(π^2 /(12))+(π^2 /(30)) =π^2 ((1/(30))−(1/(12)))=π^2 ((2/(60))−(5/(60)))=−((3π^2 )/(60))=−(π^2 /(20))

$$ \\ $$$${x}=\mathrm{sinh}\theta\Rightarrow{dx}=\mathrm{cosh}\theta\mathrm{d}\theta,\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }=\mathrm{cosh}\theta \\ $$$${x}=\mathrm{0}\Rightarrow\theta=\mathrm{0},{x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\theta=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{ln}\:\phi,\phi=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{2}{x}\right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\mathrm{d}{x}=\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \mathrm{ln}\left(\mathrm{2sinh}\:\theta\right)\mathrm{d}\theta \\ $$$$\mathrm{ln}\left(\mathrm{2}\:\mathrm{sinh}\:\theta\right)=\theta+\mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}\theta} \right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \mathrm{ln}\left(\mathrm{2}\:\mathrm{sinh}\:\theta\right)\mathrm{d}\theta=\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \left(\theta+\mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}\theta} \right)\right)\mathrm{d}\theta \\ $$$$=\left[\frac{\theta^{\mathrm{2}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{\mathrm{ln}\:\phi} +\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}\theta} \right)\mathrm{d}\theta \\ $$$${v}={e}^{−\mathrm{2}\theta} \Rightarrow\mathrm{d}\theta=−\frac{{dv}}{\mathrm{2}{v}},\theta=\mathrm{0}\Rightarrow{v}=\mathrm{1},\theta=\mathrm{ln}\:\phi\Rightarrow{v}=\phi^{−\mathrm{2}} =\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}\theta} \right)\mathrm{d}\theta=\int_{\mathrm{1}} ^{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{1}−{v}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int_{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{v}\right)}{{v}}\mathrm{d}{x} \\ $$$$\frac{\mathrm{ln}\left(\mathrm{1}−{v}\right)}{{v}}=−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}},\mid{v}\mid<\mathrm{1} \\ $$$$\int_{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}−{v}\right)}{{v}}{dv}=−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}\int_{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\mathrm{1}} {v}^{{k}−\mathrm{1}} {dv}=−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}}\left[\frac{{v}^{{k}} }{{k}}\right]_{\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}} ^{\mathrm{1}} =−\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\left(\mathrm{1}−\left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{{k}} \right) \\ $$$$=−\zeta\left(\mathrm{2}\right)+\:_{\mathrm{2}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)−,\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\:_{\mathrm{2}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{15}}−\left(\mathrm{ln}\:\:\phi\right)^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \mathrm{ln}\left(\mathrm{1}−{e}^{−\mathrm{2}\theta} \right)\mathrm{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\frac{\pi^{\mathrm{2}} }{\mathrm{15}}−\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} \right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{30}}−\frac{\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{ln}\:\phi} \mathrm{ln}\left(\mathrm{2}\:\mathrm{sinh}\:\theta\right)\mathrm{d}\theta=\frac{\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} }{\mathrm{2}}+\left(−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{30}}−\frac{\left(\mathrm{ln}\:\phi\right)^{\mathrm{2}} }{\mathrm{2}}\right)=−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\pi^{\mathrm{2}} }{\mathrm{30}} \\ $$$$=\pi^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{30}}−\frac{\mathrm{1}}{\mathrm{12}}\right)=\pi^{\mathrm{2}} \left(\frac{\mathrm{2}}{\mathrm{60}}−\frac{\mathrm{5}}{\mathrm{60}}\right)=−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{60}}=−\frac{\pi^{\mathrm{2}} }{\mathrm{20}} \\ $$

Answered by wewji12 last updated on 09/Jul/25

are you postgraduate student or professor?? i was surprised by your math skill i wonder why you were so good at math and the secret

$$\mathrm{are}\:\mathrm{you}\:\mathrm{postgraduate}\:\mathrm{student}\:\mathrm{or}\:\mathrm{professor}?? \\ $$$$\mathrm{i}\:\mathrm{was}\:\mathrm{surprised}\:\mathrm{by}\:\mathrm{your}\:\mathrm{math}\:\mathrm{skill} \\ $$$$\mathrm{i}\:\mathrm{wonder}\:\mathrm{why}\:\mathrm{you}\:\mathrm{were}\:\mathrm{so}\:\mathrm{good}\:\mathrm{at}\:\mathrm{math}\:\mathrm{and}\:\mathrm{the}\:\mathrm{secret} \\ $$

Commented by MrGaster last updated on 09/Jul/25

I'm not a postgraduate student or professor; I'm currently an eighth-grade student who has self-studied higher-level mathematics. Out of interest during the summer of 2019, I began learning on Khan Academy, targeting the AP Calculus BC exam scheduled for May of that year. After completing the AP Calculus curriculum, I moved on to study total derivatives, partial derivatives, gradients, and directional derivatives to understand the gradient descent algorithm in deep learning (the multivariable calculus on Khan Academy is taught by 3Blue1Brown!). Around September, I shifted my focus to ordinary differential equations, learning about Laplace transforms and deriving the convolution theorem.

Commented by Nicholas666 last updated on 09/Jul/25