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.
  • $\begingroup$ true, but I know that this function = 1 at the origin. $\endgroup$ – Yotam Alon Aug 13 '15 at 8:28

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*}


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.