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$$\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx$$ I have known this equals to $\frac{\pi^2}{16}$, but I cannot prove it.

metamorphy
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UntoldWind
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  • Power series look a way – YOu will not know Mar 25 '21 at 15:13
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    Hint: call it $I$. With $x=\sin t$,$$I=\int_0^{\pi/2}\ln(\sin t+\cos t)\cot tdt=\int_0^{\pi/2}\ln(\sin t+\cos t)\tan tdt,$$where th second $=$ imposes $t\mapsto\pi/2-t$. Averaging,$$I=\int_0^{\pi/2}\frac{\ln(\sin t+\cos t)dt}{\sin2t}=\frac12\int_0^{\pi/2}\frac{\ln(1+\sin2t)dt}{\sin2t}=\frac12\int_0^1dc\int_0^{\pi/2}\frac{dt}{1+c\sin2t}.$$ – J.G. Mar 25 '21 at 15:43
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    @J.G: That was a very cunning reduction! Just the complete the whole thing, With $u=\tan(t/2)$ \begin{aligned} \int^{\pi/2}_0\frac{dt}{1+c\sin 2t} &= \frac12\int^{\pi/2}_0\frac{dt}{1+c\sin t}=\frac12\int^\infty_0\frac{du}{(u+c)^2 + 1-c^2}\ &=\frac{1}{\sqrt{1-c^2}}\Big(\frac{\pi}{2}-\arctan\big(\frac{c}{\sqrt{1-c^2}}\big)\Big) \end{aligned} So $$ I=\frac{1}{4}\int^1_0\Big(\frac{\pi}{2}-\arctan\big(\tfrac{c}{\sqrt{1-c^2}}\big)\Big)\tfrac{dc}{\sqrt{1-c^2}}\stackrel{\alpha=\sin c}{=}\frac14\int^{\pi/2}_0\Big(\frac{\pi}{2}-\alpha\Big)d\alpha=\frac{\pi^2}{16}$$ – Mittens Mar 26 '21 at 16:56

1 Answers1

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Note

\begin{align} I=&\int_0^1 \frac{\ln(x+ \sqrt{1-x^2})}{x}\,dx\\ = &\int_0^1 \frac{\frac12\ln(1-x^2)+\ln(1+ \frac x{\sqrt{1-x^2}})}{x}\,dx=\frac12I_1+I_2\tag1 \end{align} where \begin{align} I_1&=\int_0^1 \frac{\ln(1-x^2)}{x}\,dx\overset{x^2\to x}=\frac12 \int_0^1 \frac{\ln(1-x)}{x}\,dx\\ &= \int_0^1 \frac{\ln(1-x)}{x}\,dx+ \int_0^1 \frac{\ln(1+x)}{x}\,dx = -\int_0^1 \frac{\ln(1+x)}{x}\,dx\\ I_2 &= \int_0^1 \frac{\ln \left(1+\frac x{\sqrt{1-x^2}}\right)}x \overset{x=\frac t{\sqrt{1+t^2}}}=\int_0^1\frac{\ln(1+t)}{t(1+t^2)}dt+ \int_1^\infty \underset{t\to1/t}{\frac{\ln(1+t)}{t(1+t^2)}dt}\\ &=\int_0^1 \frac{\ln(1+t)}t dt - \frac14 \int_0^1 \frac{\ln t}{1+t}dt =\frac54 \int_0^1 \frac{\ln(1+t)}t dt \end{align} Substitute $I_1$ and $I_2$ into (1) to obtain

$$I= \frac34 \int_0^1 \frac{\ln(1+t)}t dt =\frac34\cdot \frac{\pi^2}{12}=\frac{\pi^2}{16} $$ $\int_0^1 \frac{\ln(1+t)}t dt=\frac{\pi^2}{12}$

Quanto
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