# Does $\sum\int\arctan(n^2x)\sin(1/x)\,dx$ converge?

Does the series $$\sum_{n = 1}^\infty\int_0^{\frac{1}{n}} \arctan(n^2 x)\sin\left(\frac{1}{x}\right)dx$$ converge?

I have tried to estimate this integral, but I can't get upper estimate better than $$\frac{1}{n}$$. Can you give me a hint, please?

• Where did you get this from? That might help in figuring out what tools are expected to evaluate this. – Don Thousand May 1 at 18:25
• It is the task to prepare for the test, I don't know where it's from, but I'm quite sure that we must estimate this intergral, because Dirichlet's and Abel's test don't help – ErlGrey May 1 at 19:14

We will show that the summand $$a_{n} = \int_{0}^{1/n} \arctan(n^{2}x)\sin \left(\frac{1}{x}\right)dx = \int_n^{\infty} \arctan\left(\frac{n^{2}}{y}\right) \frac{\sin(y)}{y^{2}}dy$$ satisfies $$|a_{n}|\leq C/n^{2}$$ for some $$C>0$$, so the series converges absolutely. The key idea is that the integral is oscilating, so it cancels out a lot. We need to control it.
Choose $$N$$ such that $$2N\pi \geq n$$, so that $$N = \lceil{n/2\pi\rceil}$$. Then we can write $$a_{n} = \epsilon_{n} + \sum_{k=N}^{\infty} b_{k}$$ where $$\epsilon_{n} = \int_{n}^{2\pi N}\arctan\left(\frac{n^{2}}{y}\right) \frac{\sin(y)}{y^{2}} dy \\ b_{k} = \int_{2k\pi}^{(2k+2)\pi}\arctan\left(\frac{n^{2}}{y}\right)\frac{\sin(y)}{y}$$ From $$2\pi N \leq n + 2\pi$$, it is easy to check that $$|\epsilon_{n}| \leq C_{1}/n^{2}$$ for some $$C_{1}>0$$. For $$b_{k}$$, by splitting the integration interval as $$[2k\pi, (2k+1)\pi]$$ and $$[(2k+1)\pi, (2k+2)\pi]$$ and using $$\sin(y+\pi) = -\sin(y)$$, we get $$b_{k} = \int_{2k\pi}^{(2k+1)\pi}(f(y) - f(y+\pi)) \sin(y) dy$$ where $$f(y) = \frac{1}{y^{2}} \arctan\left(\frac{n^{2}}{y}\right).$$ By the mean value theorem, we can bound $$|f(y)-f(y+\pi)|$$ and we get $$|b_{k}| \leq \frac{1}{16k^{3}} + \frac{1}{4n^{4}k^{2}}$$ which proves $$|a_{n}| \leq O\left(\frac{1}{n^{2}}\right) + \sum_{k=N}^{\infty} \left(\frac{1}{16k^{3}} + \frac{1}{4n^{4}k^{2}}\right) \\ = O\left(\frac{1}{n^{2}}\right)$$