Explicit calculation of $\int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x$

Question

Is it possible to confirm the value of this integral using the methods of complex analysis or similar?

$$ \int_0^{\infty} \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)}\mathrm{d}x=\frac{\pi^2-9}{12} $$

Of course, one can reduce it to the definition of the polylogarithm and a $\zeta$-function, but I was looking for an explicit calculation.

Answer 1

This will require some calculations but no special functions. Making the change of variables $x = -x$ on $\mathbb R^-$, we obtain $$\int_{\mathbb R} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx = \underbrace {\int_{\mathbb R^+} \frac {2 x} {(x^2 + 1)^2 (e^{2 \pi x} - 1)} dx}_ {= I} + \int_{\mathbb R^+} \frac x {(x^2 + 1)^2} dx, \\ I = 2 \pi i \sum_{k \geq 1} \operatorname*{Res}_{x = i k} \frac x {(x^2 + 1)^2 (e^{2 \pi x} - 1)} + \frac 1 {2 (x^2 + 1)} \bigg\rvert_{x = 0}^\infty = \\ 2 \pi i \frac {4 \pi^2 + 3} {96 \pi i} - 2 \pi i \sum_{k \geq 2} \frac k {2 \pi i (k^2 - 1)^2} - \frac 1 2 = \\ \frac {4 \pi^3 - 21} {48} - \frac 1 4 \sum_{k \geq 2} \left( \frac 1 {(k - 1)^2} - \frac 1 {(k + 1)^2} \right) = \\ \frac {4 \pi^3 - 21} {48} - \frac 1 4 \left( 1 + \frac 1 4 \right) = \frac {\pi^2 - 9} {12}.$$

Answer 2

This approach is unnecessarily complicated, but quite fun:

As explained in the comments and the other answers, we may write $$ I \equiv \int \limits_0^\infty \frac{\sin(2\arctan x)}{(1+x^2)(\exp(2\pi x)-1)} \, \mathrm{d}x = \int \limits_0^\infty \frac{2 x}{(1+x^2)(\mathrm{e}^{2\pi x}-1)} \, \mathrm{d} x \, . $$ We can then use the geometric series, integration by parts and this Laplace transform to obtain \begin{align} I &= \sum \limits_{n=1}^\infty \int \limits_0^\infty \frac{2 x}{(1+x^2)^2} \mathrm{e}^{-2 \pi n x} \, \mathrm{d} x = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_0^\infty \frac{\mathrm{e}^{-2 \pi n x}}{1+x^2} \, \mathrm{d} x \right] \\ &= \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \left(\frac{\pi}{2} - \operatorname{Si}(2 \pi n)\right)\right] \, , \end{align} where $\operatorname{Si}$ is the sine integral . Plugging in its definition and integrating by parts twice more we arrive at $$ I = \sum \limits_{n=1}^\infty \left[1 - 2 \pi n \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t} \, \mathrm{d} t \right] = 2 \sum \limits_{n=1}^\infty \frac{1}{2 \pi n} \int \limits_1^\infty \frac{\sin(2 \pi n t)}{t^3} \, \mathrm{d} t \, .$$ Now the Fourier series $$ 1 - 2 \{t\} = 4 \sum \limits_{n=1}^\infty \frac{\sin(2 \pi n t)}{2 \pi n} \, , \, t \in \mathbb{R} \, , $$ and the fractional part integral $$ \int \limits_0^1 x \left\{\frac{1}{x}\right\} \, \mathrm{d} t = \int \limits_1^\infty \frac{\{t\}}{t^3} \, \mathrm{d} t = 1 - \frac{\zeta(2)}{2} = 1 - \frac{\pi^2}{12} $$ are sufficient to derive the final result $$ I = \frac{1}{2} \int \limits_1^\infty \frac{1 - 2\{t\}}{t^3} \, \mathrm{d} t = \frac{\pi^2}{12} - \frac{3}{4} = \frac{\pi^2 - 9}{12} $$ and thus prove $\pi > 3$ in a rather convoluted manner.

Answer 3

Not an answer but I believe that the form below is better for contour integration $$\int_0^\infty \frac{\sin(2\arctan(x))}{(1+x^2)(e^{2\pi x} -1)}dx=2\int_0^\infty \frac{\sin(\arctan(x))\cos(\arctan(x))}{(1+x^2)(e^{2 \pi x} -1)}dx=2\int_0^\infty \frac{x}{(1+x^2)^2 (e^{2 \pi x} -1)}dx$$ Now all that remains is to find a suitable contour that includes $[0,\infty)$, and apply residue theorem.

Answer 4

Here is an approach that first converts the integral to a double integral.

Let $$I = \int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = 2 \int_0^\infty \frac{x}{(1 + x^2)^2 (e^{2 \pi x} - 1)} \, dx.$$ Observe that $$\frac{1}{2} \int_0^\infty y e^{-y} \sin (xy) \, dy = \frac{x}{(1 + x^2)^2}.$$ Thus the integral can be rewritten as $$I = \int_0^\infty y e^{-y} \int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx \, dy, \tag1$$ after a change of order has been made.

For the inner $x$-integral \begin{align} \int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx &= \int_0^\infty \frac{e^{-2 \pi x} \sin (xy)}{1 - e^{-2 \pi x}} \, dx\\ &= \int_0^\infty \sum_{n = 0}^\infty e^{-2 \pi x} \sin (xy) \cdot e^{-2\pi n x} \, dx\\ &= \sum_{n = 1}^\infty \int_0^\infty e^{-2\pi n x} \sin (xy) \, dx, \end{align} where a shift in the index of $n \mapsto n - 1$ has been made. Now integrating by parts twice leads to $$\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx = \sum_{n = 1}^\infty \frac{y}{4 \pi^2 n^2 + y^2} = \frac{1}{2 \pi} \sum_{n = 1}^\infty \frac{\frac{y}{2\pi}}{\left (\frac{y}{2 \pi} \right )^2 + n^2}. \tag2$$

Using the well-known result of $$\pi \coth (\pi z) = \frac{1}{z} + 2 \sum_{n = 1}^\infty \frac{z}{z^2 + n^2} , \quad z \neq 0,$$ the sum in (2) can be expressed as $$\int_0^\infty \frac{\sin (xy)}{e^{2 \pi x} - 1} \, dx = \frac{1}{4} \coth \left (\frac{y}{2} \right ) - \frac{1}{2y}.$$

On returning to our integral in (1), we have \begin{align} I &= -\frac{1}{2} \int_0^\infty e^{-y} + \frac{1}{4} \int_0^\infty y e^{-y} \coth \left (\frac{y}{2} \right ) \, dy = -\frac{1}{2} + \frac{1}{4} J, \end{align} where $$J = \int_0^\infty y e^{-y} \coth \left (\frac{y}{2} \right ) \, dy.$$ Finding $J$ we have \begin{align} J &= \int_0^\infty y e^{-y} \frac{e^{y/2} + e^{-y/2}}{e^{y/2} - e^{-y/2}} \, dy\\ &= \int_0^\infty y e^{-y} \frac{1 + e^{-y}}{1 - e^{-y}} \, dy\\ &= \int_0^\infty y e^{-y} \left (1 + \frac{2 e^{-y}}{1 - e^{-y}} \right ) \, dy\\ &= \int_0^\infty y e^{-y} \, dy + 2 \int_0^\infty \frac{y e^{-2y}}{1 - e^{-y}} \, dy\\ &= 1 + 2 \sum_{n = 0}^\infty \int_0^\infty y e^{-y(n + 2)} \, dy\\ &= 1 + 2 \sum_{n = 0}^\infty \frac{1}{(n + 2)^2} \qquad \text{(by parts)}\\ &= 1 + 2 \sum_{n = 2}^\infty \frac{1}{n^2} \\ &= 1 + 2 \sum_{n = 1}^\infty \frac{1}{n^2} - 2\\ &= -1 + 2 \cdot \frac{\pi^2}{6}\\ &= -1 + \frac{\pi^2}{3}. \end{align}

So finally we have $$I = -\frac{1}{2} + \frac{1}{4} \left (-1 + \frac{\pi^2}{3} \right ),$$ or $$\int_0^\infty \frac{\sin (2 \tan^{-1} x)}{(1 + x^2) (e^{2 \pi x} - 1)} \, dx = \frac{\pi^2 - 9}{12},$$ as expected.