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This is the Monthly Problem 11939. The answer is $$ S = \frac{1}{2}\,(1 + \gamma - \log(2\pi)) + \frac{\pi^2}{72},$$ where $\gamma$ is the Euler-Mascheroni constant. My submitted solution is long. Here is a short one: http://www.mat.uniroma2.it/~tauraso/AMM/AMM11939.pdf" onclick="window.op...

Since $\sec x$ is even, we have $b_n = 0$ and $$a_n = \frac{8}{\pi}\,\int_0^{\pi/4}\sec x\cos(4nx)\,dx, \,\,(n = 0, 1, \ldots.)$$ In view of the trigonometric identity: $$\cos(4(n+1)x) - \cos(4nx) = -2\sin(2x)\sin((4n+2)x) = -4\sin x\cos x\sin((4n+2)x),$$ we find that \begin{eqnarray*} a_{n+1} - a_n...

Let the proposed integral be $I$. Notice that $$I = 2\int_0^1\log(1-x)\log(1+x)\,dx.$$ Integrating by parts yields \begin{eqnarray*} I & = & -2\int_0^1\log(1-x)\log(1+x)\,d(1-x)\\ & = & 2\int_0^1(1-x)\left(\frac{\log(1-x)}{1+ x} - \frac{\log(1+x)}{1- x}\right)\,dx\\ & = & 2\i...

We show that the series $\sum_{n=1}^\infty\,f(n)/n^2$ diverges for any bijective from $\mathbb{N}$ to $\mathbb{N}$. To see this, since $f$ is a permutation of $\mathbb{N}$, it follows that $$f(1) + f(2) + \cdots + f(n) \geq 1 + 2 + \cdots + n = \frac{n(n+1)}{2}.$$ Thus, by Abel's summation formula, ...

Notice that $$\frac{\sinh r}{\cosh n} = \frac{e^r - e^{-r}}{e^n + e^{-n}} = \frac{e^{r-n} - e^{-(r+n)}}{1 + e^{-2n}}.$$ Let $\alpha = e^{r-n}, \beta = e^{-(r+n)}$. Since $\alpha\beta = e^{-2n}$, by the formula $$\arctan\alpha - \arctan\beta = \arctan\left(\frac{\alpha - \beta}{1 + \alpha\beta}\right...

First, for $ p \geq 0$, we establish two integral formulas: \begin{eqnarray} \int_0^1\,\frac{x^p -1}{\ln x}\,dx & = &\ln(p+1).\\ \int_0^1\,\left(\frac{x^p -1}{x\ln^2 x} - \frac{p}{\ln x}\right)\,dx & =& p\ln p - p. \end{eqnarray} To this end, let $I(p) = \int_0^1\,\frac{x^p -1}{\ln x...

Another approach for extension: Recall the Euler integral $$\int_0^1\,(1-t)^pt^{q-1}\,dt = \frac{\Gamma(1+p)\Gamma(1+q)}{q\Gamma(1+p+q)}.$$ Differentiating with respect to $p$ and $q$ yields $$\int_0^1\,\ln^n(1-t)\ln^mt\frac{dt}{t} = D^n_pD^m_q\left[\frac{\Gamma(1+p)\Gamma(1+q)}{q\Gamma(1+p+q)}\righ...

Let the integral be $I$. We show that $ I = -\frac{1}{2}\zeta(4) = - \frac{\pi^4}{180}$. To see this, recall the generating function for harmonic numbers: $$\sum_{k=1}^\infty\,H_kx^k = -\,\frac{1}{1-x}\ln(1-x).$$ Integrating from $0$ to $x$ yields $$\sum_{k=1}^\infty\,\frac{H_k}{k+1}\,x^{k+1} = \fra...

Recall the well-known generating function $$\sum_{k=0}^\infty\,\binom{2k}{k}x^k = \frac{1}{\sqrt{1 -4x}}.$$ Replacing $x$ by $-x^2$ yields $$\sum_{k=0}^\infty\,(-1)^k\binom{2k}{k}x^{2k} = \frac{1}{\sqrt{1 +4x^2}}.$$ Multiplying $x^{19}$ on each side and then integrating with respect to $x$ from $0$ ...

It seems that there is a closed form for the even index. In fact, we have $$\sum_{n=1}^\infty\,\frac{H_{2n}}{(2n)\,2^n} = \frac{1}{4}\left(\ln^2\left(\frac{2 -\sqrt{2}}{2}\right)+ \ln^2\left(\frac{2 +\sqrt{2}}{2}\right)\right) + \frac{\pi^2}{48} - \frac{1}{8}\,\ln^22.$$ To this end, as we did above,...