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How can one show that

$$\int_{0}^{1}{x\ln{x}\ln(1-x^2)\over \sqrt{1-x^2}}\mathrm dx=4-{\pi^2\over 4}-\ln{4}?\tag1$$

I have tried IBP but it is seem too complicate. I am not sure what to do or how to tackle this problem. Please any help? Thank!

user153012
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The return
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  • In few words, you may enforce the substitution $x=\sin\theta$ then exploit the Fourier series of $\log\sin$ and $\log\cos$ (see here) to compute your integral through orthogonality relations. – Jack D'Aurizio Jan 22 '18 at 01:05

2 Answers2

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Enforcing a substitution of $x \mapsto \sqrt{x}$ the integral becomes $$I = \frac{1}{4} \int_0^1 \frac{\ln x \ln (1 - x)}{\sqrt{1 - x}} \, dx.$$

From the definition for Euler's beta function, namely $$\text{B}(x,y) = \int_0^1 t^{x - 1} (1 - t)^{y - 1} \, dt,$$ we observe that $$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = \int_0^1 \frac{\ln t \ln (1 - t)}{\sqrt{1 - t}} \, dt.$$

On finding the required derivatives for the Beta function we obtain $$\partial_x \partial_y \text{B}(x,y) = \text{B}(x,y) \left [\{\psi(x) - \psi(x + y)\}\{\psi(y) - \psi (x + y)\} - \psi^{(1)}(x + y) \right ].$$ Here $\psi(x)$ is the digamma function while $\psi^{(1)}(x)$ is the trigamma function or polygamma function of order one.

On taking the required limits we are left with $$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = \text{B} \left (1, \frac{1}{2} \right ) \left [\{\psi(1) - \psi(3/2)\}\{\psi (1/2) - \psi(3/2)\} - \psi^{(1)}(3/2) \right ].$$

Each of the values for the digamma and polygamma functions are well known. They are \begin{align*} \psi \left (\frac{1}{2} \right ) = -\gamma - \ln (4), & \quad \psi (1) = -\gamma,\\ \psi \left (\frac{3}{2} \right ) = 2 - \gamma - \ln (4),& \quad \psi^{(1)} \left (\frac{3}{2} \right ) = 3 \zeta (2) - 4 = \frac{\pi^2}{2} - 4. \end{align*} As $$\text{B} \left (1, \frac{1}{2} \right ) = \frac{\Gamma (1) \Gamma (1/2)}{\Gamma (3/2)} = 2,$$ we have $$\lim_{x \to 1} \lim_{y \to 1/2} \partial_x \partial_y \text{B}(x,y) = 16 - 4 \ln (4) - \pi^2,$$ yielding $$\int_0^1 \frac{x \ln x \ln (1 - x)}{\sqrt{1 - x^2}} \, dx = 4 - \ln (4) - \frac{\pi^2}{4}.$$

omegadot
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Let $x=\sin t$ and then $$I=\int_0^1 \frac{x \ln x \ln (1 - x)}{\sqrt{1 - x^2}} \, dx =2\int_0^{\pi/2}\sin t\ln\sin t\ln\cos tdt=-2\int_0^{\pi/2}\ln\sin t\ln\cos td\cos t.$$ Letting $u=\cos t$ gives \begin{eqnarray} I&=&-2\int_0^{\pi/2}\ln\sin t\ln\cos td\cos t\\ &=&\int_0^1\ln(1-u^2)\ln udu\\ &=&-\int_0^1\sum_{k=1}^\infty\frac1{k}u^{2k}\ln udu\\ &=&\sum_{k=1}^\infty\frac1{k(2k+1)^2}\\ &=&\sum_{k=1}^\infty\left[\frac{1}{k(2k+1)}-\frac{2}{(2k+1)^2}\right]. \end{eqnarray} Now using $$　\sum_{k=1}^\infty\frac{1}{k(2k+1)}=2-2\ln2 $$ and $$ \sum_{k=1}^\infty\frac{1}{(2k-1)^2}=\frac{\pi^2}{8} $$ it is easy to get $$ I=4-2\ln2-\frac{\pi^2}{4}.$$

xpaul
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