Evaluate $\lim\limits_{x\to\infty}x\left(1+\frac2\pi\int_{0}^{1}\tan^{-1}\left(t^x-t^{-x}\right)\,dt\right)$

Question

To find $\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)$ I tried the L'Hopital's rule but it cannot solve the problem. Here is my approach:

$$\lim_{x \to \infty} \frac{ 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt }{\frac{1}{x}}=\lim_{x \to \infty}\frac{\int_{0}^{1}\frac{2\ln{t}\;{\cosh{(x\ln{t})}}}{1+4\sinh^2{(x\ln{t})}}dt}{{\frac{-1}{x^2}}}=\frac{0}{0}$$ Still $\frac{0}{0}$ and second L'Hopital's won't help. Any help is appreciated.

Answer 1

Let's denote $\,I(x)=\displaystyle \int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt$

Integrating by part $$I(x)=t\tan^{-1}\left( t^x-t^{-x} \right)\Big|_0^1-x\int_{0}^{1}\frac{\left( t^x+t^{-x} \right)}{1+\left( t^x-t^{-x} \right)^2}dt$$ Making the substitution $t=e^{-s}$ $$=-x\int_0^\infty\frac{\left( e^{sx}+e^{-sx}\right)}{1+\left( e^{sx}-e^{-sx}\right)^2}e^{-s}ds\overset{t=sx}{=}-2\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}e^{-\frac{t}{x}}dt$$ With the accuracy up to exponentially small corrections at $x\gg1$, we can decompose the exponent: $\displaystyle e^{-\frac{t}{x}}=1-\frac{t}{x}+\frac{t^2}{2x^2}+O\Big(\frac{1}{x^3}\Big)$ $$I(x)=-2\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}dt+\frac{2}{x}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}t\,dt+O\Big(\frac{1}{x^2}\Big)$$ $$=-\frac{\pi}{2}+\frac{2}{x}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t\,dt-\frac{1}{x^2}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t^2\,dt+O\Big(\frac{1}{x^3}\Big)$$ and, therefore, the desired limit $$\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)=\frac{4}{\pi}\int_0^\infty\frac{\cosh t}{1+4\sinh^2t}\,t\,dt=0.77749...$$

(it can also be presented in the form $\displaystyle -\,\frac{2}{\pi}\int_0^1\frac{1+t^2}{1-t^2+t^4}\ln t\,dt\,$).

The numeric evaluation with WolframAlpha confirms the answer.

$\bf{Update}$

As @170231 mentioned, the last integral has a closed form: $$ J=\int_0^1\frac{1+t^2}{1-t^2+t^4}\ln t\,dt=-\,\frac{4}{3}G$$ and the desired limit is $$\boxed{\,\,\lim_{x \to \infty} x\left( 1+\frac{2}{\pi}\int_{0}^{1}\tan^{-1}\left( t^x-t^{-x} \right)dt \right)=\frac{8G}{3\pi}=0.77749...\,\,}$$ Indeed, $$J=\int_0^1\frac{(1+t^2)^2}{1+t^6}\ln t\,dt=$$ $$\int_0^1(1-t^6+t^{12}-...)\ln t\,dt+2\int_0^1(t^2-t^8+t^{14}-...)\ln t\,dt$$ $$+\int_0^1(t^4-t^{10}+t^{14}-...)\ln t\,dt$$ $$=-1+\frac{1}{7^2}-\frac{1}{13^2}+..+2\Big(-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{15^2}-..\Big)-\frac{1}{5^2}+\frac{1}{11^2}-\frac{1}{17^2}+...$$ Denoting the Catalan' constant $\displaystyle G=1-\frac{1}{3^2}+\frac{1}{5^2}-...$ $$J=-1+\frac{1}{3^3}-\frac{1}{5^2}+\frac{1}{7^2}-\frac{1}{9^2}+ ... -\frac{2}{9}G-\frac{1}{3^2}+\frac{1}{9^2}-\frac{1}{15^2}+...$$ $$=-G-\frac{2}{9}G-\frac{1}{9}G=-\,\frac{4}{3}G$$ It is interesting to note that the second term of the asymptotics $I(x)$ can also be found in a closed form: $$I(x)=I_0+\frac{1}{x}I_1+\frac{1}{x^2}I_2\,+...$$ where $$I_2=-\int_0^\infty\frac{t^2\cosh t}{1+4\sinh^2t}\,dt=-\frac{1}{2}\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}\int_{-\infty}^\infty\frac{e^{-\beta t}\cosh t}{1+4\sinh^2t}dt=-\frac{1}{2}\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}J(\beta)$$ The last integral can be evaluated by means of integration along a rectangular contour in the complex plane: $$J(\beta)=\frac{\pi}{2}\,\frac{\cos\frac{\beta\pi}{3}}{\cos\frac{\beta\pi}{2}};\quad\frac{\partial^2}{\partial \beta^2}\,\bigg|_{\beta=0}J(\beta)=\frac{5\pi^3}{72}$$ and $$I(x)=-\frac{\pi}{2}+\frac{4G}{3}\,\frac{1}{x}-\frac{5\pi^3}{144}\,\frac{1}{x^2}+O\Big(\frac{1}{x^3}\Big)$$

Answer 2

Partial evaluation of @Svyatoslav's log integral:

$$\begin{align*} J &= \int_0^1 \frac{1+t^2}{1-t^2+t^4} \ln(t) \, dt \\[1ex] &= -\frac{e^{i\pi/3}}{1-e^{i\pi/3}} \int_0^1 \frac{\ln(t)}{t^2-e^{i\pi/3}} \, dt - \frac{e^{-i\pi/3}}{1-e^{-i\pi/3}} \int_0^1 \frac{\ln(t)}{t^2-e^{-i\pi/3}} \, dt \tag{1} \\[1ex] &= e^{i\pi/3} \int_0^1 \frac{\ln(t)}{1-e^{-i\pi/3}t^2} \, dt + e^{-i\pi/3} \int_0^1 \frac{\ln(t)}{1-e^{i\pi/3}t^2} \, dt \\[1ex] &= \sum_{n=0}^\infty \left( e^{-i(n-1)\pi/3} + e^{i(n-1)\pi/3}\right) \int_0^1 t^{2n} \ln(t) \, dt \tag{2} \\[1ex] &= -2 \sum_{n=0}^\infty \frac{\cos\left(\frac{(n-1)\pi}3\right)}{(2n+1)^2} \tag{3} \\[1ex] &= - \sum_{n=0}^\infty \frac{\cos\left(\frac{n\pi}3\right)+ \sqrt3 \sin\left(\frac{n\pi}3\right)}{(2n+1)^2} \end{align*}$$

The remaining sum can be expressed in terms of hypergeometric functions, such that

$$J = e^{i2\pi/3} {}_{3}F_{2} \left(\left.\begin{array}{c|c}\frac12,\frac12,1\\\frac32,\frac32\end{array}\right\vert e^{i\pi/3}\right) + e^{-i2\pi/3} {}_{3}F_{2} \left(\left.\begin{array}{c|c}\frac12,\frac12,1\\\frac32,\frac32\end{array}\right\vert e^{-i\pi/3}\right)$$

but it's not immediately clear to me how to simplify this...

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• $(1)$ : partial fractions
• $(2)$ : $\displaystyle \frac1{1-x}=\sum_{n\ge0}x^n$
• $(3)$ : integrate by parts and simplify

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Update: A more complete evaluation of the integral

Taking the sum we get after integrating in line $(2)$, preserving the exponential form, we get

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

where $z={e^{i\pi/3}}$.

We have for $|x|<1$, using the fundamental theorem of calculus,

$$\begin{align*} f(x) &= \sum_{n=0}^\infty \frac{x^{2n+1}}{(2n+1)^2} \\[2ex] \implies f'(x) &= \sum_{n=0}^\infty \frac{x^{2n}}{2n+1} = \frac{\operatorname{artanh}(x)}{x} \\[2ex] \implies f(x) &= \int_0^x \frac{\operatorname{artanh}(y)}{y} \, dy \\[1ex] &= \frac12 \int_0^x \frac{\ln(1+y) - \ln(1-y)}{y} \, dy \\[1ex] &= \operatorname{Li}_2(x) - \frac14 \operatorname{Li}_2(x^2) \\[2ex] \implies \sum_{n=0}^\infty \frac{x^{n-1}}{(2n+1)^2} &= \frac{f(\sqrt x)}{x\sqrt x} \end{align*}$$

and hence another form for the integral in terms of the dilogarithm,

$$J = - \frac{f(\sqrt z)}{z\sqrt z} - \frac{f(\sqrt{\bar z})}{\bar z \sqrt{\bar z}}$$

where I assume a branch cut along the negative real axis for the square root. This reduces further but I'm not as well-versed in the dilogarithm identities as I could be. Mathematica uses the same branch, and evaluates either sum with $z$ or $\bar z$ as $\dfrac23G \pm i \dfrac{\pi^2}{12}$, hence $J=-\dfrac43G$.

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In[]:= f[E^(I \[Pi]/6)]/(E^(I \[Pi]/3) E^(I \[Pi]/6)) // FunctionExpand // Expand
Out[]= (2 Catalan)/3 - (I [Pi]^2)/12