# Convergence of an improper integral with trig function

$$\int_0^{ + \infty } {\left( {\frac{x}{{1 + {x^6}{{\sin^2 x}}}}} \right)dx}$$

I'd like your help with see why does?? I tried to compare it to other functions and to change the variables, but it didn't work for me.

Thanks a lot!

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what do you mean by why does ??. fyi, Mathematica says this integral does not converge. – Nasser Nov 27 '12 at 4:55

We can divide up the ray $[0,\infty)$ into subintervals $[(n-\frac{1}{2})\pi,(n+\frac{1}{2})\pi]$, $n=1,2,\ldots$ (there a little piece $[0,\pi/2]$ left over, which we may ignore). On each subinterval, we compute $$\begin{eqnarray*} \int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{x}{1+x^6\sin^2 x}\,dx&\leq& (n+\frac{1}{2})\pi\int_{(n-\frac{1}{2})\pi}^{(n+\frac{1}{2})\pi}\frac{1}{1+x^6\sin^2 x}\,dx\\ &\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+(n-\frac{1}{2})^6\pi^2\sin^2 u}\,du\\ &\leq&(n+\frac{1}{2})\pi\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\ &\leq&(n+\frac{1}{2})\pi\int_{-\infty}^{\infty}\frac{1}{1+4(n-\frac{1}{2})^6u^2}\,du\\ &=&\frac{\pi^2(n+\frac{1}{2})}{2(n-\frac{1}{2})^3} \end{eqnarray*}$$
These integrals are decaying like $\frac{1}{n^2}$, so the sum over all $n$ converges.
+1. There might be little error though. How did you go from the second line to third line? You could directly integrate $$\int_{-\pi/2}^{\pi/2} \dfrac{du}{1+a^2 \sin^2(u)} = \dfrac{\pi}{\sqrt{1+a^2}}$$ – user17762 Nov 27 '12 at 6:50
@Marvis: on $[-\frac\pi2,\frac\pi2]$, $4u^2\le\pi^2\sin^2(u)$. – robjohn Nov 27 '12 at 13:39