Show that $\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2))$

Question

Show that $$\int_{0}^{2e}{\ln{(x^2)}\ln{(4e-x)}\over \sqrt{x(4e-x)}} dx=\pi(1-\zeta(2)).$$

My try: let$u=\ln{x}$ then $x\mathrm du=\mathrm dx$ and $$2\int{ue^{u/2}\over \sqrt{4e-e^u}}\mathrm du$$ What next?

Answer 1

By letting $x=2e(1-\cos(t))$, we have that $$\begin{align} I:=\int_{0}^{2e}\frac{\ln(x^2)\ln(4e-x)}{\sqrt{x(4e-x)}}dx =2\int_{0}^{\pi/2}\ln(2e(1-\cos(t)))\ln(2e(1+\cos(t)))dt. \end{align}$$ Now by using the Fourier series given by robjohn in this answer, we have that $$\ln(2e(1+\cos(t)))=\ln(2)+1+\ln(1+\cos(t)) =1-2\sum_{k=1}^\infty(-1)^{k}\frac{\cos(kt)}{k}$$ and $$\ln(2e(1-\cos(t)))=\ln(2)+1+\ln(1-\cos(t)) =1-2\sum_{k=1}^\infty\frac{\cos(kt)}{k}.$$ Hence, after taking the product and the integration over $(0,\pi/2]$, we get $$I=2\cdot\frac{\pi}{2}-8\cdot\frac{\pi}{4}\sum_{k=1}^\infty\frac{(-1)^k}{k^2} =\pi-2\pi\cdot\frac{\pi^2}{12}=\pi(1-\zeta(2)).$$

Answer 2

Let $$I = \int_0^{2e} \frac{\ln (x^2) \ln (4e - x)}{\sqrt{x(4e - x)}} \, dx = 2 \int_0^{2e} \frac{\ln (x) \ln (4e - x)}{\sqrt{x(4e - x)}} \, dx.$$ Under the transformation $x \mapsto 4ex$ we have \begin{align*} I &= 2 \int_0^{1/2} \frac{\ln (4ex) \ln [4e(1 - x)]}{\sqrt{x (1 - x)}} \, dx\\ &= 2 \ln^2 (4e) \int^{1/2}_0 \frac{dx}{\sqrt{x (1 - x)}} + 2 \ln (4e) \left \{\int^{1/2}_0 \frac{\ln x}{\sqrt{x (1 - x)}} \, dx + \int^{1/2}_0 \frac{\ln (1 - x)}{\sqrt{x (1 - x)}} \, dx \right \} + 2 \int^{1/2}_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx\\ &= 2 \ln^2 (4e) I_1 + 2\ln (4e) I_2 + 2 I_3. \end{align*}

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Finding $I_1$

Consider

$$\int^1_0 \frac{dx}{\sqrt{x (1 - x)}} = \int^{1/2}_0 \frac{dx}{\sqrt{x (1 - x)}} + \int^1_{1/2} \frac{dx}{\sqrt{x (1 - x)}}.$$ If in the rightmost integral we let $x \mapsto 1 - x$ one has \begin{align*} \int^1_0 \frac{dx}{\sqrt{x (1 - x)}} &= \int^{1/2}_0 \frac{dx}{\sqrt{x (1 - x)}} + \int^{1/2}_0 \frac{dx}{\sqrt{x (1 - x)}}\\ &= 2 \int^{1/2}_0 \frac{dx}{\sqrt{x (1 - x)}}. \end{align*} Thus $$I_1 = \frac{1}{2} \int^1_0 \frac{dx}{\sqrt{x (1 - x)}}.$$ The above integral can be found in terms of a Beta function. Here $$I_1 = \frac{1}{2} \int^1_0 x^{-1/2} (1 - x)^{-1/2} \, dx = \frac{1}{2} \int^1_0 x^{\frac{1}{2} - 1} (1 - x)^{\frac{1}{2} - 1} \, dx = \frac{1}{2} \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) = \frac{\pi}{2}.$$

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Finding $I_2$

As $$I_2 = \int^{1/2}_0 \frac{\ln x}{\sqrt{x (1 - x)}} \, dx + \int^{1/2}_0 \frac{\ln (1 - x)}{\sqrt{x (1 - x)}} \, dx,$$ if in the rightmost integral we let $x \mapsto 1 - x$ one obtains $$I_2 = \int^{1/2}_0 \frac{\ln x}{\sqrt{x (1 - x)}} \, dx + \int_{1/2}^1 \frac{\ln x}{\sqrt{x (1 - x)}} \, dx = \int^1_0 \frac{\ln x}{\sqrt{x (1 - x)}} \, dx.$$ This integral can be found using the derivative of a beta function as follows. Since $$\text{B}(x,y) = \int^1_0 t^{x - 1} (1 - t)^{y - 1} \, dt,$$ we see that $$I_2 = \partial_x \text{B}(x,y) \Big{|}_{x,y = 1/2}.$$

Now as $$\partial_x \text{B}(x,y) = \text{B}(x,y) \left \{\psi(x) - \psi (x + y) \right \},$$ where $\psi (x)$ is the digamma function, we have $$I_2 = \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) \left \{\psi \left (\frac{1}{2} \right ) - \psi (1) \right \} = -\pi \ln (4).$$ Here we have used the following well-known values for the digamma function of $\psi (1/2) = - \gamma - \ln (4)$ and $\psi (1) = -\gamma$ where $\gamma$ corresponds to the Euler-Mascheroni constant.

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Finding $I_3$

Consider $$\int^1_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx = \int^{1/2}_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx + \int_{1/2}^1 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx.$$ If in the rightmost integral we again let $x \mapsto 1 - x$, then \begin{align*} \int^1_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx &= \int^{1/2}_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx + \int^{1/2}_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx\\ &= 2 \int^{1/2}_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx, \end{align*} or $$I_3 = \frac{1}{2} \int^1_0 \frac{\ln x \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx.$$

$I_3$ can again be found using a method that employs the derivative of the beta function. Here $$I_3 = \frac{1}{2} \partial_y \partial_x \text{B} (x,y) \Big{|}_{x,y = 1/2}.$$

Now as $$\partial_y \partial_x \text{B}(x,y) = \text{B}(x,y) \left \{[\psi (x) - \psi (x + y)][\psi (y) - \psi (x + y)] - \psi^{(1)}(x + y) \right \},$$ where $\psi^{(m)}(x)$ is the polygamma function of order $m$, this gives $$\partial_y \partial_x \text{B}(x,y) \Big{|}_{x,y = 1/2} = \text{B} \left (\frac{1}{2}, \frac{1}{2} \right ) \left \{ \left [\psi \left (\frac{1}{2} \right ) - \psi (1) \right ]^2 - \psi^{(1)} (1) \right \} = \pi (\ln^2 (4) - \zeta (2)).$$ Here the value $\psi^{(1)} (1) = \zeta (2)$ has been used. Thus $$I_3 = \frac{\pi}{2} (\ln^2 (4) - \zeta (2)).$$

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So returning to the original integral we have $$I = 2 \ln^2 (4e) \cdot \frac{\pi}{2} - 2 \ln (4e) \cdot \pi \ln (4) + 2 \cdot \frac{\pi}{2} (\ln^2 (4) - \zeta (2)),$$ or $$\int^{2e}_0 \frac{\ln (x^2) \ln (1 - x)}{\sqrt{x (1 - x)}} \, dx = \pi (1 - \zeta (2)),$$ as required.