Christmas integral

Let the integral in the numerator be

$I_1 = \int_{0}^{\infty} \frac{e^{-t}\sin{t}}{t}dt$

This integral will be solved using the concept of Laplace transforms as such:

Consider the Laplace transform of $\sin{t}$ :

$L\left[\sin{t}\right]= \int_{0}^{\infty} e^{-st}\sin(t)dt$

By referring to a table of Laplace transforms, this standard integral evaluates to:

$\int_{0}^{\infty} e^{-st}\sin(t)dt = \frac{1}{1+s^2}$

Generally for Laplace transforms:

$L\left[f(t)\right] = \int_{0}^{\infty} e^{-st}f(t)dt = F(s)$

Now, using the table of Laplace transforms, one can use the identity:

$L\left[\frac{f(t)}{t}\right] = \int_{0}^{\infty} \frac{e^{-st}f(t)}{t}dt = \int_{s}^{\infty}F(p)dp$

Using this identity for the Laplace transform of a sine function, one gets:

$L\left[\frac{\sin{t}}{t}\right] = \int_{0}^{\infty} \frac{e^{-st}\sin{t}}{t}dt = \int_{s}^{\infty}\frac{1}{1+p^2}dp$

Taking $s=1$ gives:

$\boxed{I_1=\int_{0}^{\infty} \frac{e^{-t}\sin{t}}{t}dt = \int_{1}^{\infty}\frac{1}{1+p^2}dp=\frac{\pi}{4}}$

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Let the integral in the denominator be:

$I_2 = \int_{0}^{\infty} \frac{x}{e^x-1}dx=\int_{0}^{\infty} \frac{e^{-x}x}{1-e^{-x}}dx$

The function $e^{-x}$ is less than unity in the interval $(0,\infty)$ . Therefore, the following term can be written as an infinite geometric series as such:

$\frac{e^{-x}}{1-e^{-x}} = e^{-x}+e^{-2x}+e^{-3x}+\dots$

Replacing this in the integral gives:

$I_2 = \int_{0}^{\infty} \left(e^{-x}+e^{-2x}+e^{-3x}+\dots\right)xdx = \sum_{k=1}^{\infty}\left(\int_{0}^{\infty}xe^{-kx}\right)dx$

Now, by introducing a change of variables $kx=z$ , and after some simplification, the sum of integrals transforms to the sum:

$I_2 = \left(\sum_{k=1}^{\infty}\frac{1}{k^2}\right)\int_{0}^{\infty}ze^{-z}dz$

The integral:

$\int_{0}^{\infty}ze^{-z}dz=1$

and the following sum refers to the Basel problem: $\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}$

Therefore:

$\boxed{I_2 = \int_{0}^{\infty} \frac{x}{e^x-1}dx = \frac{\pi^2}{6}}$

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The required expression to be evaluated is:

$\boxed{\frac{8I_1^2}{I_2}=3}$

Relevant references just a google search away:

• The Reimann-Zeta function
• Laplace Transforms
• The Basel problem

See, $\int_{0}^\infty \frac{sin(x)e^{-x}}{x} dx$

$=\int_{0}^\infty (1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+....)e^{-x} dx$

$=\sum_{0}^{\infty} \frac{(-1)^k \int_{0}^\infty x^{(2k+1)-1} e^{-x} dx}{(2k+1)!}$

$=\sum_{0}^\infty (-1)^{k}\frac{\Gamma(2k+1)}{(2k+1)!}$

$=\Gamma(1)-\frac{\Gamma(3)}{3!}+\frac{\Gamma(5)}{5!}-\frac{\Gamma(7)}{7!}-....$

$=1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$ = $\int_{0}^1 \frac{1}{1+x^2} dx=\frac{π}{4}$

Again we know $\int_{0}^\infty \frac{x}{e^x-1} dx=\zeta(2)=\frac{π^2}{6}$ .

Then the desired result is $\frac{8\frac{π^2}{16}}{\frac{π^2}{6}}=3$ .