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Recall the generating function for harmonic numbers: $$\sum_{n=1}^\infty\,H_nx^n = - \frac{1}{1-x}\,\ln(1-x).$$ Integrating this yields $$\sum_{n=1}^\infty\,\frac{H_n}{n+1}x^{n+1} = \frac{1}{2}\,\ln^2(1-x).$$ This can be rewritten as $$\sum_{n=1}^\infty\,\frac{H_n}{n}x^{n} = \frac{1}{2}\,\ln^2(1-x) ...

Let the proposed integral be $I$. We show that $I = \pi^4/15$. Indeed, \begin{eqnarray*} I & = & \int_0^\infty\,\frac{x^3e^{-x}}{1 - e^{-x}}\,dx\\ & = & \int_0^\infty\,\sum_{n=0}^\infty\,x^3e^{- (n+1)x}\,dx\\ & = & \sum_{n=0}^\infty\,\int_0^\infty\,x^3e^{- (n+1)x}\,dx\\ &...

For example, you can find the following formula

$$\int_0^\infty\,x\sin(ax^2)\sin(2bx)\,dx = \frac{b}{2a}\sqrt\frac{\pi}{2a}\left(\cos\frac{b^2}{a} + \sin\frac{b^2}{a}\right)$$

in Gradshteyn & Ryzhik's Table of Integrals, Series, and Products as Formula 3.851.1, on page 500 (5th edition).

It seems that the value of the series is $\frac{3-e}{2\sqrt{e}}$. To this end, recall the exponentinal generating function of $\lfloor n!/e\rfloor$ (due to Mark Hoeij, 2011) $$\sum_{n=0}^\infty\,\frac{1}{n!}\,\left\lfloor\frac{n!}{e}\right\rfloor\,x^n = -\frac{1}{2}\left(e^x + \frac{x+1}{x-1}\,e^{-x...

Some Remarks: By the substitution $x = t^2$, the proposed integral becomes $$2\,\int_0^\infty\,t\sin(t^2)\sin t\,dt.$$ In 1851, Cauchy obtained $$ \int_0^\infty\,\sin(t^2)\cos(pt)\,dt = \frac{1}{2}\,\sqrt{\frac{\pi}{2}}\,\left(\cos\frac{p^2}{4} - \sin\frac{p^2}{4}\right).$$ He then differentiated u...

Here is the third way. Without loss of generality, we assume that $\alpha \geq 0$. Let $$f(\alpha) = \int_0^\infty\,\frac{\cos\alpha x}{1 + x^2}\,dx$$ Applying Leibniz's rule gives $$f'(\alpha) = -\int_0^\infty\,\frac{x\sin\alpha x}{1 + x^2}.$$ Appealing to $$\frac{\sin\alpha x}{x(1 + x^2)} = \frac{...

Notice that \begin{eqnarray*} \int_0^\infty\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx & = & \int_0^1\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx + \int_1^\infty\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx\\ & = & 2\,\int_0^1\,\frac{\ln^2x}{1 + 2x\cos\theta + x^2}\,dx, \end{eqnarray*} ...

\begin{eqnarray*} \sum_{n=1}^\infty\,\zeta(n+1)x^n & = & \sum_{n=1}^\infty\left(\sum_{k=1}^\infty\,\frac{1}{k^{n+1}}\right)x^n \\ & = & \sum_{k=1}^\infty\,\frac{1}{k}\left(\sum_{n=1}^\infty\,\frac{x^n}{k^n}\right)\\ & = & \sum_{k=1}^\infty\,\frac{1}{k}\cdot\frac{x}{k-x}\\ &am...

Under the additional condition $|\alpha| < 1$, we show the derivatives of both sides are equal. Indeed, term by term differentiating with respect to $x$ yields \begin{eqnarray*} \left( \sum_{k=1}^\infty\,\frac{\alpha^k}{k}\,\sin kx \right)' & = & \sum_{k=1}^\infty\,\alpha^k\,\cos kx\\ & ...

Just add one remark. This problem is a special case of the Simpson's dissection method: If $f(x) = \sum_{n=0}^\infty\,a_nx^n$, then $$\sum_{n=0}^\infty\,a_{kn+m}x^{kn+m} = \frac{1}{k}\,\sum_{j=0}^{k-1}\,\omega^{-jm}f(\omega^jx),$$ where $\omega = e^{2\pi i/k}$ is a primitive $k$th root of unity. App...