Calculation of an indefinite integral containing $\sin(x)$ and $x$

Question

I have to calculate this simple integral: $$J(x)=\int_0^x\frac{1}{\sin(t)-t}dt$$ Despite its apparent simplicity, I can't figure it. Is there some method to calculate it? Thanks.

Answer 1

For small positive $t$ (say $0 < t < 1)$, we have $- \frac16 t^3 < \sin t - t < -\frac16 t^3 + \frac{1}{120}t^5 < - \frac{19}{120} t^3$. Thus for $0 < \delta < \varepsilon < 1$, we have

$$-6\int_\delta^\varepsilon \frac{dt}{t^3} > \int_\delta^\varepsilon \frac{dt}{\sin t - t} > - \frac{120}{19} \int_\delta^\varepsilon \frac{dt}{t^3}.$$

Since $\lim\limits_{\delta \searrow 0} \int_\delta^\varepsilon \frac{dt}{t^3} = +\infty$, we have a non-integrable singularity in $0$. If we want to interpret

$$J(x) = \int_0^x \frac{dt}{\sin t-t}$$

at all, we must interpret it as $J(x) = -\infty$ for $x > 0$ (and $J(x) = +\infty$ for $x < 0$).