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$$ \sum_{n=1}^\infty \int_n^{n+1} \frac{\sin\pi x}{x^p+1} \,\mathrm d x, \qquad 0<p\leq 1 $$

The reference answer only additionally hints to use the first mean value theorem for definite integrals.

According to the hint, my thoughts are as follows, $$ \begin{align*}& \int_n^{n+1} \frac{\sin\pi x}{x^p+1} \,\mathrm d x\\ <& \left|\sin\bigl( (n+\frac{1}{2}) \pi \bigr)\right|\, \int_n^{n+1} \frac{1}{x^p+1} \,\mathrm d x\\ =& \int_n^{n+1} \frac{1}{x^p+1} \,\mathrm d x\\ <& \int_n^{n+1} \frac{1}{x^p} \,\mathrm d x =\left.\frac{x^{1 - p}}{1 - p}\right\vert_n^{n+1}\\ =& \frac{{(n + 1)}^{1 - p} - n^{1 - p}}{1 - p} \end{align*} $$

I know I'm obviously on the wrong path...

mamath
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1 Answers1

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Notice that $\sin(\pi x)$ doesn't change its sign on the interval $[n,n+1]$ since its zeros are $x=k\in\mathbb{Z}$. By the mean value theorem there's a $c_n\in[n,n+1]$ such that $$\int_n^{n+1}\frac{\sin(\pi x)}{x^p+1}\,dx=\frac{1}{(c_n)^p+1}\int_n^{n+1}\sin(\pi x)\,dx $$ For the last integral, we have $$\int_n^{n+1}\sin(\pi x)\,dx =-\frac{1}{\pi}\cos(\pi x)\Bigg\vert_n^{n+1}\\ =-\frac{1}{\pi}\left(\cos((n+1)\pi)-\cos(n\pi)\right)=\frac{2\cdot(-1)^n}{\pi}$$ so you have to establish the convergence of $$\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^n}{(c_n)^p+1} $$ which follows from the alternating series test, and the divergence of $$\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{(c_n)^p+1} $$ which follows from a comparison with the divergent series $\frac{2}{\pi}\sum_{n=1}^{\infty}\frac{1}{(n+1)^p+1}=\frac{2}{\pi}\sum_{n=2}^{\infty}\frac{1}{n^p+1} $.

bjorn93
  • 6,787