Evaluating $\int_0^1 \int_0^1 \frac{\frac{\pi}{4} dx dy}{\sin(\frac{\pi}{4}(1+x+y))}$

Question

I'm hoping to evaluate the definite integral $$I = \int_0^1 \int_0^1 \frac{\frac{\pi}{4} dx dy}{\sin(\frac{\pi}{4}(1+x+y))}$$

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This integral arose in an attempt to evaluate $\int_{0}^\infty \left( \frac{x-1}{\log(x)} \right)^2 \frac{dx}{1+x^4}$; these integrals are equal and I will accept an evaluation of either.

It is straightforward to directly perform the integral on one of the coordinates via the antiderivative $\int \csc(x) dx = -\log\left( |\cot(x) + \csc(x)|\right)$. I suspect that there may not be a nice evaluation of the integral on the remaining coordinate, but I'm still hopeful.

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Spelling out the integration over the coordinate $x$, we have

$$ \begin{split} I &= \int_0^1 dy [ -\log(\cot(\frac{\pi}{4}(1+x+y)+\csc(\frac{\pi}{4}(1+x+y)) )] \bigg|_{x=0}^{x=1} \\ &= \int_{0}^1 dy \log \left(\frac{\cot(\frac{\pi}{4}(1+y)) +\csc(\frac{\pi}{4}(1+y)) }{\cot(\frac{\pi}{4}(2+y)) +\csc(\frac{\pi}{4}(2+y))} \right) \end{split}$$

Answer 1

Make the substitution $u = \frac{x+y}2$ and $v=\frac{x-y}2$. $$I=\frac{\pi}{2}\int_{0}^{\frac{1}{2}}\int_{-u}^u \frac{1}{\sin(\frac\pi4 (1+2u))}dvdu+\frac{\pi}{2}\int_{\frac12}^{1}\int_{u-1}^{1-u} \frac{1}{\sin(\frac\pi4 (1+2u))}dvdu$$ Performing the innermost integral is now trivial. $$I=\pi\int_{0}^{\frac{1}{2}} \frac{u}{\sin(\frac\pi4 (1+2u))}du+\pi\int_{\frac12}^{1} \frac{1-u}{\sin(\frac\pi4 (1+2u))}du$$ Now, we substitute $u\to 1-t$ on the second integral. $$I=\pi\int_{0}^{\frac{1}{2}} \frac{u}{\sin(\frac\pi4 (1+2u))}du+\pi\int_{0}^{\frac12} \frac{t}{\sin(\frac\pi4 (3-2t))}dt$$ By using the trig identity $\sin(\pi-\theta)=\sin(\theta)$, we can see the integrands are the same.

$$I=2\pi\int_{0}^{\frac{1}{2}} \frac{u}{\sin(\frac\pi4 (1+2u))}du$$ Now substitute $\theta =\frac\pi4 (1+2u)$ and split the resulitng integrals. $$I=\frac{8}{\pi}\int_{\frac\pi4}^{\frac\pi2}\frac{\theta}{\sin(\theta)}d\theta-2\int_{\frac\pi4}^{\frac\pi2}\frac{1}{\sin(\theta)}d\theta$$ The second term is elementary, equaling $2\ln(\sqrt2 -1)$. However, I doubt the first term has an elementary form. But there is something nice we can do to it. Perform the half angle substitution: $z=\tan(\frac\theta2)$. $$I=\frac{16}{\pi}\int_{\sqrt{2}-1}^{1}\frac{\arctan(z)}{z}dz +2\ln(\sqrt2 -1)$$

The antiderivative of $\frac{\arctan(z)}{z}$ is the inverse tangent integral and is known to be transcendental as it can be written in terms of the dilogarithm. In particular, it has a nice power series. $$\mathrm{T}(z) =\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{(2k+1)^2} = z-\frac{z^3}{3^2}+\frac{z^5}{5^2}-\frac{z^7}{7^2}\dots$$ Notice that $\mathrm{T}(1)$ is Catalan's constant $G$. Therfore, $$I = \frac{16}{\pi}G + 2\ln(\sqrt2 -1) - \sum_{k=0}^\infty \frac{(-1)^k(\sqrt2 -1)^{2k+1}}{(2k+1)^2}$$

Answer 2

Here is a approach. Let $I$ denote the integral. Then substituting $x = e^t$, we get

\begin{align*} I = \int_{0}^{\infty} \left( \frac{x-1}{\log x}\right)^2 \frac{\mathrm{d}x}{x^4 + 1} = \int_{-\infty}^{\infty} \frac{e^t(e^t - 1)^2}{t^2(e^{4t} + 1)} \, \mathrm{d}t = \int_{-\infty}^{\infty} \frac{f(t)}{t^2} \, \mathrm{d}t \end{align*}

where

$$ f(z) = \frac{2\sinh^2(z/2)}{\cosh(2z)}. $$

Note that $f(z)$ is a $2\pi i$-periodic meromorphic function on $\mathbb{C}$ that converges to $0$ uniformly in $\operatorname{Re}{z} \to \pm \infty$. So by considering a CCW-oriented rectangular contour $\mathcal{C}_{R,N}$ with the four corners $\pm R$, $\pm R + 2N \pi i$ for $R > 0$ and $N \in \mathbb{N}$ and then letting $R \to \infty$ and $N \to \infty$ so that $Ne^{-R} \to 0$, we get

\begin{align*} I &= \lim_{\substack{R,N \to \infty \\ Ne^{-R} \to 0}} \oint_{\mathcal{C}_{R,N}} \frac{f(z)}{z^2} \, \mathrm{d}z \\ &= 2\pi i \sum_{k=1,3,5,\ldots} \underset{z=k\pi i/4}{\mathrm{Res}} \frac{f(z)}{z^2} \qquad \text{as $N\to\infty$} \\ &= \frac{8}{\pi} \biggl( \frac{2-\sqrt{2}}{1^2} + \frac{-2-\sqrt{2}}{3^2} + \frac{2+\sqrt{2}}{5^2} + \frac{-2+\sqrt{2}}{7^2} \\ &\hspace{2em} + \frac{2-\sqrt{2}}{9^2} + \frac{-2-\sqrt{2}}{11^2} + \frac{2+\sqrt{2}}{13^2} + \frac{-2+\sqrt{2}}{15^2} + \cdots \biggr) \\ &= \frac{16}{\pi} \biggl( \frac{5}{4} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^2} - \sum_{k=1}^{\infty} \frac{\sin(\pi k/4)}{k^2} \biggr) \\ &= \frac{16}{\pi} \biggl( \frac{5}{4} G - \operatorname{Im}(\operatorname{Li}_2(e^{i\pi/4}) \biggr). \end{align*}

Here,

$$ G = \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{(2k-1)^2} \qquad\text{and}\qquad \operatorname{Li}_2(z) = \sum_{k=1}^{\infty} \frac{z^k}{k^2} $$

are Catalan's constant and the dilogarithm function, respectively. As to my knowledge, neither $G$ nor $\operatorname{Im}(\operatorname{Li}_2(e^{i\pi/4}))$ has known elementary closed forms.

Answer 3

Feynman’s Trick brings us a nice answer

Let’s consider the parametrised integral $$ I(a)=\int_0^{\infty}\left(\frac{x^a-1}{\ln x}\right)^2 \frac{d x}{1+x^4} $$ Then differentiating once and twice yields $$ I^{\prime}(a)=2 \int_0^{\infty} \frac{x^{2 a}-x^a}{\ln x\left(1+x^4\right)} d x $$ and $$ \begin{aligned} I^{\prime \prime}(a) & =2 \int_0^{\infty} \frac{2 x^{2 a}-x^a}{1+x^4} d x \\ & =\frac{\pi}{2}\left[2 \csc \frac{(2 a+1) \pi}{4}-\csc \frac{(a+1) \pi}{4}\right] \end{aligned} $$ Integrating back gives $$ \begin{aligned} I^{\prime}(a) & =I^{\prime}(a)-I^{\prime}(0)=\int_0^a I^{\prime \prime}(u) d u \\ &=\frac{\pi}{2} \int_0^a \left[2 \csc \frac{(2 u+1) \pi}{4}-\csc \frac{(u+1) \pi}{4}\right] d u\\&= -2 \ln \left(\csc \frac{(2 a+1) \pi}{4}+\cot \frac{(2 a+1) \pi}{4}\right)+2 \ln \left(\csc \frac{(a+1) \pi}{4}+\cot\frac{(a+1 )\pi}{4}\right) \end{aligned} $$ Integrating back again brings us $$ \begin{aligned} &\int_0^{\infty}\left(\frac{x-1}{\ln x}\right)^2 \frac{d x}{1+x^4} \\=&I(1)-I(0) \\ = & \int_0^1\left[-2 \ln \left(\csc \frac{(2 a+1) \pi}{4}+\cot \frac{(2 a+1) \pi}{4}\right)+2 \ln \left(\csc \frac{(a+1) \pi}{4}+\cot \frac{(a+1) \pi}{4}\right) \right]da\\ = & -\frac{4}{\pi} \underbrace{ \int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \ln (\csc x+\cot x) d x}_{J} + \underbrace{ \frac{8}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \ln (\csc x+\cot x)dx }_{K} \\ \end{aligned} $$

$$\therefore \boxed{I=\frac{4}{\pi}\left[3 G-4 \Im\left(\operatorname{Li_2}\left(-e^{-\frac{\pi}{4} i}\right)\right)\right] }$$

where $$J \stackrel{x\mapsto x-\frac{\pi}{2}}{=} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \ln (\sec x-\tan x) d x=0 $$ and

$$ \begin{aligned} K= & \int_0^{\frac{\pi}{2}} \ln (1+\cos x) d x-\int_0^{\frac{\pi}{4}} \ln (1+\cos x) d x -\int_0^{\frac{\pi}{2}} \ln (\sin x) d x+\int_0^{\frac{\pi}{4}} \ln (\sin x) d x \\ = & \left(2 G-\frac{\pi}{2} \ln 2\right)-\left(2 \Im\left[\operatorname{Li_2}\left(-e^{-\frac{\pi}{4} i}\right)\right]-\frac{\pi}{4} \ln 2\right) +\frac{\pi}{2} \ln 2-\frac{1}{4}(2 G+\pi \ln 2)\\=& \frac{3}{2} G-\Im\left[\operatorname{Li}_2\left(-e^{-\frac{\pi}{4} i}\right)\right] \end{aligned}\\ $$

with my post .

Answer 4

Here is yet another method, starting from where I began in the question.

$$\begin{split} I &= \int_0^1 dy [ -\log(\cot(\frac{\pi}{4}(1+x+y)+\csc(\frac{\pi}{4}(1+x+y)) )] \bigg|_{x=0}^{x=1} \\ &= \int_{0}^1 dy \log \left(\frac{\cot(\frac{\pi}{4}(1+y)) +\csc(\frac{\pi}{4}(1+y)) }{\cot(\frac{\pi}{4}(2+y)) +\csc(\frac{\pi}{4}(2+y))} \right)\\ &=\bigg[\int_0^1 dy \log(1+\cos(\frac{\pi}{4}(1+y))) - \int_0^1 dy \log(1+\cos(\frac{\pi}{4}(2+y)))\\& - \int_0^1 \log(\sin(\frac{\pi}{4}(1+y)) + \int_0^1 \log(\sin(\frac{\pi}{4}(2+y)) \bigg] \end{split} $$

In the following, I'll try a sequence of substitutions to simplify the integrals. There may be a slightly more compact choice of substitutions. Depending on the arguments of the trig functions, first I substitute $\frac{\pi}{4}(1+y) \to y$, or I do $\frac{\pi}{4}y \to y$ and use the identity $\cos(\frac{\pi}{2} + x) = -\sin(x)$.

$$ \begin{split} I &= \frac{4}{\pi}\bigg[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log(1+\cos(y)) - \int_0^{\frac{\pi}{4}} dy \log(1-\sin(y))\\& - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \log(\sin(y)) + \int_0^{\frac{\pi}{4}} \log(\cos(y)) \bigg]\end{split} $$

Next I try the substitution $y \to \frac{\pi}{2} - y$ in the second and third integrals. $$ \begin{split} I &= \frac{4}{\pi}\bigg[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log(1+\cos(y)) - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log(1-\cos(y))\\& - \int_0^{\frac{\pi}{4}} \log(\cos(y)) + \int_0^{\frac{\pi}{4}} \log(\cos(y)) \bigg] \\&= \frac{4}{\pi}\bigg[\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log(1+\cos(y)) - \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log(1-\cos(y)) \bigg]\end{split} $$

We can continue to massage this expression into ones we can evaluate.

$$ \begin{split} I &= \frac{4}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log \bigg(\frac{1+\cos(y)}{1-\cos(y)}\bigg) \\&= \frac{4}{\pi}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dy \log \bigg((\cot(\frac{x}{2}))^2 \bigg) \\&= \frac{16}{\pi}\int_{\frac{\pi}{8}}^{\frac{\pi}{4}} dy \log \big(\cot(x) \big) \\&= \frac{16}{\pi}\bigg[ \int_{0}^{\frac{\pi}{4}} dy \log \big(\cot(x) \big) + \int_{0}^{\frac{\pi}{8}} dy \log \big(\tan(x) \big) \bigg] \\&= \frac{16}{\pi}\bigg[ G + \int_{0}^{\frac{\pi}{8}} dy \log \big(\tan(x) \big) \bigg] \end{split} $$

$\int_{0}^{\frac{\pi}{4}} dy \log \big(\cot(x) \big) $ is Catalan's constant $G$, while there are a few representations of $\int_{0}^{\frac{\pi}{8}} dy \log \big(\tan(x) \big)$ in the answers here, such as $\int_{0}^{\frac{\pi}{8}} dy \log \big(\tan(x) \big)=\frac{\pi}{8}\ln\big(\sqrt{2}-1\big)-\Im{\left(\text{Li}_2\left(i\big(\sqrt{2}-1\big)\right)\right)}$.

Answer 5

After struggling with the integral in the post, I found that

$$\int_0^{\infty}\left(\frac{x-1}{\ln x}\right)^2 \frac{d x}{1+x^4} = \frac{4}{\pi}\left[3 G-4 \Im\left(\operatorname{Li_2}\left(-e^{-\frac{\pi}{4} i}\right)\right)\right] ,$$

then I started to generalise the integral as $$I_n=\int_0^{\infty}\left(\frac{x-1}{\ln x}\right)^2 \frac{d x}{1+x^n}, $$ where $n\ge 3$.

Replacing the power $4$ in my solution by $n$, we have

$$ I_n=-\frac{n}{\pi} \int_{\frac{\pi}{n}}^{\frac{3 \pi}{n}} \ln (\cos x+\cot x) d x+\frac{2 n}{\pi} \int_{\frac{\pi}{n}}^{\frac{2 \pi}{n}} \ln (\cos x+\cot x) d x $$ Now we need to evaluate the integral using Clausen Function.

$$ \begin{aligned} \int_0^t \ln (\csc x+\cot x) d x = & \int_0^t \ln \left(\frac{1+\cos x}{\sin x}\right) d x \\ = & \int_0^t \ln \left(\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) d x \\ = & -\int_0^t \ln \left(\tan \frac{x}{2}\right) d x \\ = & -2\left[-\frac{1}{2} \operatorname{Cl_2}(t)-\frac{1}{2} \operatorname{Cl_2}(\pi-t)\right] \\ = & \operatorname{Cl_2}(t)+ \operatorname{Cl_2}(\pi-t) \end{aligned} $$

Plugging back the limits yields

$$ \begin{aligned} I_n&=-\frac{n}{\pi} \int_{\frac{\pi}{n}}^{\frac{3 \pi}{n}} \ln (\cos x+\cot x) d x+\frac{2 n}{\pi} \int_{\frac{\pi}{n}}^{\frac{2 \pi}{n}} \ln (\csc x+\cot x) d x \\ &=-\frac{n}{\pi}\left[\operatorname{Cl_2}\left(\frac{3 \pi}{n}\right)+ \operatorname{Cl_2}\left(\pi-\frac{3 \pi}{n}\right)-\operatorname{Cl_2}\left(\frac{\pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{\pi}{n}\right)\right] \\ & \quad +\frac{2 n}{\pi}\left[\operatorname{Cl_2}\left(\frac{2 \pi}{n}\right)+ \operatorname{Cl_2}\left(\pi-\frac{2 \pi}{n}\right)-\operatorname{Cl_2}\left(\frac{\pi}{n}\right)-\left.C\right|_2\left(\pi-\frac{\pi}{n}\right)\right] \\ &= \frac{n}{\pi}\left[2\operatorname{Cl_2}\left(\frac{2 \pi}{n}\right)+2 \operatorname{Cl_2}\left(\pi-\frac{2 \pi}{n}\right)-\operatorname{Cl_2}\left(\frac{3 \pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{3 \pi}{n}\right)\right. \\ &\left.\quad-\operatorname{Cl_2}\left(\frac{\pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{\pi}{n}\right)\right] \end{aligned} $$

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For examples,

$$I_3=\frac{3}{\pi} \Im\left[\operatorname{Li}_2\left((-1)^{\frac{2}{3}}\right)+\operatorname{Li}_2\left((-1)^{\frac{1}{3}}\right)\right] $$

$$ I_4=\frac{8}{\pi}\left[2 G-\Im\left(\operatorname{Li_2}((-1)^{\frac{3}{4}})+\operatorname{Li_2}((-1)^{\frac{1}{4}})\right) \right] $$

checked by WA.

Answer 6

Feynman’s Trick brings us a nice answer

Let’s consider the parametrised integral $$ I(a)=\int_0^{\infty}\left(\frac{x^a-1}{\ln x}\right)^2 \frac{d x}{1+x^n}, $$ where $n\ge 3$. Then differentiating once and twice yields $$ I^{\prime}(a)=2 \int_0^{\infty} \frac{x^{2 a}-x^a}{\ln x\left(1+x^n\right)} d x $$ and $$ \begin{aligned} I^{\prime \prime}(a) & =2 \int_0^{\infty} \frac{2 x^{2 a}-x^a}{1+x^n} d x \\ & =\frac{\pi}{2}\left[2 \csc \frac{(2 a+1) \pi}{n}-\csc \frac{(a+1) \pi}{n}\right] \end{aligned} $$ Integrating back gives $$ \begin{aligned} I^{\prime}(a) & =I^{\prime}(a)-I^{\prime}(0)=\int_0^a I^{\prime \prime}(u) d u \\ &=\frac{\pi}{2} \int_0^a \left[2 \csc \frac{(2 u+1) \pi}{n}-\csc \frac{(u+1) \pi}{n}\right] d u\\&= -2 \ln \left(\csc \frac{(2 a+1) \pi}{n}+\cot \frac{(2 a+1) \pi}{n}\right)+2 \ln \left(\csc \frac{(a+1) \pi}{n}+\cot\frac{(a+1 )\pi}{n}\right) \end{aligned} $$ Integrating back once more brings us $$ \begin{aligned} &\int_0^{\infty}\left(\frac{x-1}{\ln x}\right)^2 \frac{d x}{1+x^n} \\=&I(1)-I(0) \\ = & \int_0^1\left[-2 \ln \left(\csc \frac{(2 a+1) \pi}{n}+\cot \frac{(2 a+1) \pi}{n}\right)+2 \ln \left(\csc \frac{(a+1) \pi}{n}+\cot \frac{(a+1) \pi}{n}\right) \right]da\\ = & -\frac{n}{\pi} \underbrace{ \int_{\frac{\pi}{n}}^{\frac{3 \pi}{n}} \ln (\csc x+\cot x) d x}_{J} + \underbrace{ \frac{2n}{\pi}\int_{\frac{\pi}{n}}^{\frac{2\pi}{n}} \ln (\csc x+\cot x)dx }_{K} \\ \end{aligned}$$

Now we need to evaluate the integrals $J$ and $K$ using Clausen Function.

$$ \begin{aligned} \int_0^t \ln (\csc x+\cot x) d x = & \int_0^t \ln \left(\frac{1+\cos x}{\sin x}\right) d x \\ = & \int_0^t \ln \left(\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right) d x \\ = & -\int_0^t \ln \left(\tan \frac{x}{2}\right) d x \\ = & -2\left[-\frac{1}{2} \operatorname{Cl_2}(t)-\frac{1}{2} \operatorname{Cl_2}(\pi-t)\right] \\ = & \operatorname{Cl_2}(t)+ \operatorname{Cl_2}(\pi-t) \end{aligned} $$

Plugging back the limits yields

$$ \begin{aligned} I_n&=-\frac{n}{\pi} \int_{\frac{\pi}{n}}^{\frac{3 \pi}{n}} \ln (\cos x+\cot x) d x+\frac{2 n}{\pi} \int_{\frac{\pi}{n}}^{\frac{2 \pi}{n}} \ln (\csc x+\cot x) d x \\ &=-\frac{n}{\pi}\left[\operatorname{Cl_2}\left(\frac{3 \pi}{n}\right)+ \operatorname{Cl_2}\left(\pi-\frac{3 \pi}{n}\right)-\operatorname{Cl_2}\left(\frac{\pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{\pi}{n}\right)\right] \\ & \quad +\frac{2 n}{\pi}\left[\operatorname{Cl_2}\left(\frac{2 \pi}{n}\right)+ \operatorname{Cl_2}\left(\pi-\frac{2 \pi}{n}\right)-\operatorname{Cl_2}\left(\frac{\pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{\pi}{n}\right)\right] \\ \end{aligned} $$ Simplification concludes that

$$\boxed{\quad \int_0^{\infty}\left(\frac{x-1}{\ln x}\right)^2 \frac{d x}{1+x^n}\\= \frac{n}{\pi}\left[2\operatorname{Cl_2}\left(\frac{2 \pi}{n}\right)+2 \operatorname{Cl_2}\left(\pi-\frac{2 \pi}{n}\right)-\operatorname{Cl_2}\left(\frac{3 \pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{3 \pi}{n}\right)\right. \\ \left.\quad-\operatorname{Cl_2}\left(\frac{\pi}{n}\right)-\operatorname{Cl_2}\left(\pi-\frac{\pi}{n}\right)\right] }$$

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For examples,

$$I_3=\frac{3}{\pi} \Im\left[\operatorname{Li}_2\left((-1)^{\frac{2}{3}}\right)+\operatorname{Li}_2\left((-1)^{\frac{1}{3}}\right)\right] $$

$$ I_4=\frac{8}{\pi}\left[2 G-\Im\left(\operatorname{Li_2}((-1)^{\frac{3}{4}})+\operatorname{Li_2}((-1)^{\frac{1}{4}})\right) \right] $$

checked by WA.