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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?

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1 Answer 1

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Cornel says that

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

Hint: Your integral parametrization above is good, and great if you cleverly use it (think of differentiating twice). Also, you may improve slightly in terms of simplicity in the part with Change of variables.

End of story {at least a result in (Almost) Impossible Integrals, Sums, and Series (2019) may be found useful in the solving process}

A note: I have no idea who Martin.s is - it happened the last four answers went to his/her/their questions.

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  • $\begingroup$ Martin.s, enjoy the closed form! :-) (at any rate, again I probably won't be around for a long time) $\endgroup$ Commented May 19 at 7:38
  • $\begingroup$ Thanks a lot for the answer, but are my calculations correct? Also, you mentioned that the integral is evaluated by Cornel, but I'm not able to find the integral in either of the books. $\endgroup$
    – Martin.s
    Commented May 19 at 8:26

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