# how to evaluate $\int_{0}^{1}\int_{0}^{1} \frac{2-x-y}{\sqrt{xy}(1+xy) \log_{\pi}^{2}{xy}} \, dx \, dy$

How to evaluate: \begin{align*} &\int_{0}^{1}\int_{0}^{1} \frac{2-x-y}{(\sqrt{xy}+\sqrt{x^{3} y^{3}}) \left[ \log_{\pi}^{2}{x} + 4\log_{\pi}{\sqrt{x}} \log_{\pi}{y} + \log_{\pi}^{2}{y} \right]} \, dx \, dy \end{align*}

### My attempt(almost a complete solution)

$$I := \int_{0}^{1}\int_{0}^{1} \frac{2-x-y}{(\sqrt{xy}+\sqrt{x^{3} y^{3}}) \left[ \log_{\pi}^{2}{x} + 4\log_{\pi}{\sqrt{x}} \log_{\pi}{y} + \log_{\pi}^{2}{y} \right]} \, dx \, dy$$

$$= \int_{0}^{1}\int_{0}^{1} \frac{2-x-y}{(\sqrt{xy}+\sqrt{x^{3} y^{3}}) \left[ \log_{\pi}^{2}{x} + 2\log_{\pi}{x} \log_{\pi}{y} + \log_{\pi}^{2}{y} \right]} \, dx \, dy$$

$$= \int_{0}^{1}\int_{0}^{1} \frac{2-x-y}{\sqrt{xy}(1+xy) \log_{\pi}^{2}{xy}} \, dx \, dy$$

$$= \ln^{2}{\pi} \int_{0}^{1}\int_{0}^{1} \frac{2-x-y}{\sqrt{xy}(1+xy) \log_{\pi}^{2}{xy}} \, dx \, dy$$

Change of variables: $$\begin{cases} u = xy \\ v = y \end{cases} \quad , \quad \frac{\partial (u, v)}{\partial (x, y)} = \begin{vmatrix} y & x \\ 0 & 1 \end{vmatrix} = y = v$$

Thus: $$\begin{cases} x = \frac{u}{v} \\ y = v \end{cases} \quad \text{and} \quad \left| \frac{\partial (x, y)}{\partial (u, v)} \right| = \frac{1}{v}$$

So: \begin{align*} I &= \ln^{2}{\pi} \iint_{D} \frac{2 - \frac{u}{v} - v}{\sqrt{u}(1+u) \ln^{2}{u}} \cdot \frac{1}{v} \, du \, dv \end{align*}

Where: $$D = \left\{ (u, v) \, | \, 0 \leq u \leq v \leq 1 \right\}$$

Therefore: $$I = \ln^{2}{\pi} \int_{0}^{1} \frac{1}{\sqrt{u}(1+u) \ln^{2}{u}} \left[ \int_{u}^{1} \left( \frac{2}{v} - \frac{u}{v^{2}} - 1 \right) \, dv \right] \, du$$

$$= \ln^{2}{\pi} \int_{0}^{1} \frac{1}{\sqrt{u}(1+u) \ln^{2}{u}} \left[ \left( 2 \ln{v} + \frac{u}{v} - v \right) \bigg|_{u}^{1} \right] \, du$$

$$= 2 \ln^{2}{\pi} \int_{0}^{1} \frac{u-1-\ln{u}}{\sqrt{u}(1+u) \ln^{2}{u}} \, du \quad \left( \text{let } u = e^{-t} \right)$$

$$= 2 \ln^{2}{\pi} \int_{0}^{+\infty} \frac{e^{-\frac{t}{2}}}{1+e^{-t}} \cdot \frac{e^{-t} - 1 + t}{t^{2}} \, dt$$

Let: $$J(\lambda) := \int_{0}^{+\infty} \frac{e^{-\frac{t}{2}}}{1+e^{-t}} \cdot \frac{e^{-\lambda t} - 1 + \lambda t}{t^{2}} \, dt \quad (\lambda \geq 0)$$

Then: \begin{align*} J'(\lambda) &= \int_{0}^{+\infty} \frac{e^{-\frac{t}{2}}}{1+e^{-t}} \cdot \frac{1 - e^{-\lambda t}}{t} \, dt \end{align*}

And: $$J''(\lambda) = \int_{0}^{+\infty} \frac{e^{-\frac{t}{2}}}{1+e^{-t}} \cdot e^{-\lambda t} \, dt$$

$$= \int_{0}^{+\infty} \sum_{n=0}^{+\infty} (-1)^{n} \cdot e^{-(\lambda + n + \frac{1}{2}) t} \, dt$$

$$= \sum_{n=0}^{+\infty} (-1)^{n} \int_{0}^{+\infty} e^{-(\lambda + n + \frac{1}{2}) t} \, dt$$

$$= \sum_{n=0}^{+\infty} \frac{(-1)^{n}}{\lambda + n + \frac{1}{2}}$$

Given: $$J'(0) = 0 \Rightarrow J'(\lambda) = \sum_{n=0}^{+\infty} (-1)^{n} \cdot \ln{\left(1 + \frac{2\lambda}{2n+1}\right)}$$

And: $$J(0) = 0 \Rightarrow J(\lambda) = \sum_{n=0}^{+\infty} (-1)^{n} \cdot \left[ \left( \lambda + \frac{2n+1}{2} \right) \ln{\left(1 + \frac{2\lambda}{2n+1}\right)} - \lambda \right]$$

So: \begin{align*} I &= 2 \ln^{2}{\pi} \cdot J(1) \\ &= 2 \ln^{2}{\pi} \cdot \sum_{n=0}^{+\infty} (-1)^{n} \cdot \left[ \frac{2n+3}{2} \ln{\left(1 + \frac{2}{2n+1}\right)} - 1 \right] \end{align*}

I don't know how to evaluate the last sum, and can someone check if my proof is correct?

$$\int_{0}^{\infty} \frac{e^{-\frac{t}{2}}}{1+e^{-t}} \cdot \frac{e^{-t} - 1 + t}{t^{2}} \textrm{d}t=2\frac{ G}{\pi }-\frac{1}{2}-\frac{3}{2}\log (2)-\log (\pi)+2\log\left( \Gamma\left(\frac{1}{4}\right)\right).$$