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Does the integral $\int_0^{\infty} {1 \over {1+x^2 \cdot \sin^2 x}}$ converge?

What I tried:

  • substituting $x \cdot \sin x$ with $t$ and transforming into arctan of something.
  • tried to find a multiplication of useful functions in order to do integration by parts.
  • I thought about trying to change the integral into a sum if integrals from $k \pi$ to $(k+1) \pi$ but that didn't really give me anything new (except that it produced a series that converges which says nothing for the sum). also, none of the convergence tests that I know gave any conclusive result.
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  • $\begingroup$ true, but I know that this function = 1 at the origin. $\endgroup$ – Yotam Alon Aug 13 '15 at 8:28
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The following lemma is useful for our aim:

Lemma. For any $a > 0$ we have $$ \int_{0}^{\pi} \frac{dx}{1+a^2\sin^2 x} = \frac{\pi}{\sqrt{1+a^2}}. $$

Proof of this lemma is straightforward:

\begin{align*} \int_{0}^{\pi} \frac{dx}{1+a^2\sin^2 x} &= 2 \int_{0}^{\pi/2} \frac{\sec^2 x}{1+(1+a^2)\tan^2 x} \, dx \\ &= 2 \int_{0}^{\infty} \frac{dt}{1+(1+a^2)t^2} \, dt \qquad (t = \tan x) \\ &= \frac{\pi}{\sqrt{1+a^2}}. \end{align*}

Now we have

\begin{align*} \int_{0}^{\infty} \frac{dx}{1+x^2 \sin^2 x} & \geq \sum_{k=1}^{\infty} \int_{(k-1)\pi}^{k\pi} \frac{dx}{1+(k\pi)^2 \sin^2 x} \\ &= \sum_{k=1}^{\infty} \frac{\pi}{\sqrt{1+k^2\pi^2}} \\ &= \infty. \end{align*}

Remarks.

  1. I finally recalled that I posted a solution to an analogous problem. You may check this answer. Depending on your flavor, it may look easier to you.

  2. This argument also shows that $$ \int_{0}^{R} \frac{dx}{1+x^2 \sin^2 x} = \log R + \mathcal{O}(1) \quad \text{as } R \to \infty. $$

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