Evaluate $\int_{0}^{1} \frac{\ln x}{(1+8x^2)\sqrt{1-x^2}} dx$

Question

Stuck in by parts ( or any other method is appreciated ): $$ \int_{0}^{1}\frac{\ln\left(x\right)} {\left(1 + 8x^{2}\right)\sqrt{1 - x^{2}}}\,{\rm d}x $$

• To evaluate this integral, I have tried many different ideas but none seems to get me to any result.
• I have tried $$ x = \sin\left(t\right):\ \mbox{integral of}\quad \frac{\ln\left(\sin\left(x\right)\right)}{1 + 8 \sin^{2}\left(x\right)}, $$ then by parts taking $\ln\left(\sin\left(x\right)\right)$ as first function, but I am getting stuck in integration by parts of $$ \cot\left(x\right)\arctan\left(3\tan\left(x\right)\right) $$ Please help.

Answer 1

$$I=\int_0^1\frac{\ln(x)}{(1 + 8 x^2) \sqrt{1 - x^2}}\,dx$$

Put $x=\cos\theta$ $$I=\int_0^\frac\pi2 \frac{\ln(\cos\theta)}{1+8\cos^2\theta}\,d\theta$$ Put $t=\tan\theta$

$$I=\int_0^{\infty}\frac{\ln\left(\frac{1}{\sqrt{1+t^2}}\right)}{1+\left(\frac{8}{{1+t^2}}\right)}\frac{1}{1+t^2}\,dt=\frac{-1}{2}\int_0^{\infty}\frac{\ln(1+t^2)}{9+t^2}\,dt$$

$$\boxed{\int_{0}^{\infty} \frac{\ln(a^{2}+x^{2})}{b^{2}+x^{2}} \, dx = \frac{\pi}{b} \, \ln(a+b)}$$

The above result is from here

Set $a=1,b=3$ in the above formula;

$$I=-\frac{\pi}{3} \, \ln(2)\approx -0.725$$

Which also agrees with this

Answer 2

Starting with OP’s substitution, $x=\sin \theta$, then $$ I=\int_0^{\frac{\pi}{2}} \frac{\ln (\sin \theta)}{1+8 \sin ^2 \theta} d \theta $$ Let $t=\cot \theta$, we have $$ \begin{aligned} I & =\int_{\infty}^0 \frac{\ln \left(\frac{1}{\sqrt{1+t^2}}\right)}{1+\frac{t}{1+t^2}} \cdot \frac{-d t}{1+t^2} \\ & =-\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{9+t^2} d t \\ & =-\frac{1}{4} \int_{-\infty}^{\infty} \frac{\ln \left(1+t^2\right)}{9+t^2} d t \\ & =-\frac{1}{4}\cdot 2 \Re \int_{-\infty}^{\infty} \frac{\ln (t+i)}{9+t^2} d t \\ & =-\frac{1}{4} \cdot 2 \Re\left[2 \pi i \cdot \lim_{t\rightarrow 3i}(t-3 i) \frac{\ln(t+i)}{t^2+9 }\right] \\ & =-\frac{1}{2} \Re\left[2 \pi i \cdot \frac{\ln (4 i)}{6 i}\right] \\ & =-\frac{\pi}{3} \ln 2 \end{aligned} $$

Answer 3

Let $t=\frac{\sqrt {1-x^2}}{x} $, then $x^2=\frac{1}{1+t^2}$ and $$ \begin{aligned}\int_0^1 \frac{\ln x}{\left(1+8 x^2\right) \sqrt{1-x^2}} d x = & \int_{\infty}^0 \frac{\ln \left(\frac{1}{\sqrt{1+t^2}}\right)}{\left(1+\frac{8}{1+t^2}\right)\left(\frac{t}{\sqrt{1+t^2}}\right)} \cdot\left(-\frac{t}{\left(1+t^2\right)^{\frac 32}}\right) dt\\ = & -\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+t^2\right)}{9+t^2} d t \\ = & -\frac{1}{2} \cdot \frac{2 \pi\ln 2}{3} \quad \text { (as above) } \\ = & -\frac{\pi\ln 2}{3} \end{aligned} $$

Answer 4

$$\begin{align*} I &= \int_0^1 \frac{\log x}{\left(1+8x^2\right) \sqrt{1-x^2}} \, dx \\ &= \int_0^1 \int_x^1 \cdots \, dy \, dx = - \int_0^1 \int_0^y \frac{dx\,dy}{y\left(1+8x^2\right)\sqrt{1-x^2}} \\ &= -\frac13 \int_0^1 \arctan\frac{\sqrt{1-y^2}}{3y} \cdot \frac{dy}y & (*) \\ &= -3 \int_0^\infty \frac{\arctan z}{z\left(z^2+9\right)} \, dz & (*) \\ &= \int_0^\infty \int_0^1 \cdots \, dw \, dz = -3 \int_0^1 \int_0^\infty \frac{dz\,dw}{\left(z^2+9\right)\left(z^2w^2+1\right)} \\ &= \frac{3\pi}2 \int_0^1 \left(\frac w{1-9w^2} - \frac1{3-27w^2}\right) \, dw \\ &= \frac{3\pi}2 \left(-\frac{\log2}6 - \frac{\log2}{18}\right) = \boxed{-\frac\pi3\log2} \end{align*}$$

where the integrals labeled $(*)$ are obtained by leveraging an Euler substitution, e.g.

$$\int\frac{dx}{\left(1+8x^2\right)\sqrt{1-x^2}} \stackrel{t=\tfrac x{\sqrt{1-x^2}}}=\int\frac{dt}{1+9t^2}=\frac13\arctan(3t)+C$$

Answer 5

Let's make some calculations that will help us out later,

$$J=\int \frac{1}{(1 + 8 x^2) \sqrt{1 - x^2}}\,dx$$

$$= \frac13\int \frac{3}{(1 + 8 x^2) \sqrt{1 - x^2}}\,dx$$

$$=\frac13\int \frac{3(1-x^2)+3x^2}{(1 - x^2+9x^2) \sqrt{1 - x^2}}\,dx$$

$$=\frac13\int \frac{3\sqrt{1-x^2}+\frac{3x^2}{\sqrt{1-x^2}}}{\left(1 +\frac{9x^2}{1-x^2}\right)({1 - x^2})}\,dx$$

$$=\frac13\int \frac{1}{1 +\left(\frac{3x}{\sqrt{1-x^2}}\right)^2}\,d\left({\frac{3x}{\sqrt{1-x^2}}}\right)$$

$$=\frac13 \arctan \left(\frac{3x}{\sqrt{1-x^2}}\right)$$

---

$$\int_0^1 \frac{\ln(x)}{(1 + 8 x^2) \sqrt{1 - x^2}}\,dx=-\frac13\int_0^1 \frac1x \arctan \left(\frac{3x}{\sqrt{1-x^2}}\right) \,dx$$

Coming back to $I$, After Integrating by parts, we get,

$$I=-\frac13\int_0^1 \frac1x \arctan \left(\frac{3x}{\sqrt{1-x^2}}\right) \,dx$$

Consider, $$I(a)=-\frac13\int_0^1 \frac1x \arctan \left(\frac{3ax}{\sqrt{1-x^2}}\right) \,dx$$

$$\frac{\partial}{\partial a}I(a)=-\int_0^1 \frac{\sqrt{1-x^2}}{1-x^2+9a^2x^2}\,dx$$

$$=\frac{1}{9a^2-1}\int_0^1 \frac{\sqrt{1-x^2}(1-9a^2)}{1-x^2+9a^2x^2}\,dx$$

$$=\frac{1}{9a^2-1}\int_0^1 \frac{(1-x^2+9a^2x^2)-9a^2}{(1-x^2+9a^2 x^2)(\sqrt{1-x^2})}\,dx$$

$$=\frac{1}{9a^2-1}\int_0^1 \frac{1}{\sqrt{1-x^2}}\,dx+\frac{1}{1-9a^2}\int_0^1 \frac{9a^2}{(1-x^2+9a^2 x^2)(\sqrt{1-x^2})}\,dx$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{1}{1-9a^2}\int_0^1 \frac{9a^2}{(1-x^2+9a^2 x^2)(\sqrt{1-x^2})}\,dx$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{1}{1-9a^2}\int_0^1 \frac{9a^2+9a^2x^2-9a^2x^2}{(1-x^2)\left(1+\frac{9a^2 x^2}{1-x^2}\right)(\sqrt{1-x^2})}\,dx$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{1}{1-9a^2}\int_0^1 \frac{9a^2(1-x^2)+9a^2x^2}{(1-x^2)\left(1+\frac{9a^2 x^2}{1-x^2}\right)(\sqrt{1-x^2})}\,dx$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{1}{1-9a^2}\int_0^1 \frac{9a^2\sqrt{1-x^2}+\frac{9a^2x^2}{\sqrt{1-x^2}}}{(1-x^2)\left(1+\left(\frac{3ax}{\sqrt{1-x^2}}\right)^2\right)}\,dx$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{3a}{1-9a^2}\int_0^1 \frac{3a\sqrt{1-x^2}+(2x)(3ax)\left(\frac{1}{2\sqrt{1-x^2}}\right)}{1+\left(\frac{3ax}{\sqrt{1-x^2}}\right)^2}\,dx$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{3a}{1-9a^2}\int_0^1 \frac{1}{1+\left(\frac{3ax}{\sqrt{1-x^2}}\right)^2}\,d\left(\frac{3ax}{\sqrt{1-x^2}} \right)$$

$$=\frac{\pi}{2(9a^2-1)}+\frac{3a\pi}{2(1-9a^2)}=\frac{(3a-1)\pi}{2(1-9a^2)}$$

$$\int_0^1 dI(a)=\frac{\pi}{2}\int_0^1\frac{(3a-1)}{1-9a^2}\,da$$

$$I(1)\underset{I(0)=0}=\frac{\pi}{2}\int_0^1\frac{(3a-1)}{(1-9a^2)}\,da=-\frac{\pi}{2}\int_0^1\frac{(3a-1)}{(9a^2-1)}\,da$$

$$I(1)=-\frac{\pi}{2}\int_0^1 \frac{1}{3a+1}\,da=-\frac{\pi}{3}\ln(2)$$

Therefore,

$$\color{red}{\int_0^1 \frac{\ln(x)}{(1 + 8 x^2) \sqrt{1 - x^2}}\,dx=-\frac{\pi}{3}\ln(2)}$$