Another limit involving an integral

Question

How to find the following limit (if exists ) : $\lim _{x \to \infty} \dfrac 1 x \int_0^x \dfrac {t}{1+x^2 \cos^2 t}dt$ ?

This looks quite similar to the previous To find a limit involving integral , but the huge difference is , there is apparently no closed form formula for the indefinite integral $\int \dfrac {t}{1+x^2 \cos^2 t}dt$ , where there was a closed form formula for the previous one , which made its evaluation slightly easy . Please help .

Answer 1

I am using the same method as in this answer you linked. Let

$$ F(x)=\frac{1}{x}\int_0^x \frac{t}{1+x^2\cos^2{t}}\mathrm{d}t. $$

If $n$ is a positive integer, we have

$$ F(n\pi)=\frac{1}{n\pi}\sum_{k=0}^{n-1}\int_{k\pi}^{(k+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t, $$

so we can control the $t$ in the integrand :

$$\frac{1}{n}\sum_{k=0}^{n-1}k\int_{k\pi}^{(k+1)\pi} \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}} \leq F(n\pi)\leq \frac{1}{n}\sum_{k=0}^{n-1}(k+1)\int_{k\pi}^{(k+1)\pi} \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}}. $$

Using the identity $\frac{1}{\pi}\int_0^\pi\frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}}=\frac{1}{\sqrt{1+n^2\pi^2}}$, that yields

$$\frac{1}{n}\frac{\pi}{\sqrt{1+n^2\pi^2}}\sum_{k=0}^{n-1}k\leq F(n\pi)\leq \frac{1}{n}\frac{\pi}{\sqrt{1+n^2\pi^2}}\sum_{k=0}^{n-1}(k+1), $$

hence

$$F(n\pi) \to \frac{1}{2}. $$

Now when $n\pi \leq x < (n+1)\pi$,

$$ \begin{align} F(x) &\leq \frac{1}{n\pi}\int_0^{(n+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t \\ &=F(n\pi)+\frac{1}{n\pi}\int_{n\pi}^{(n+1)\pi} \frac{t}{1+(n\pi)^2\cos^2{t}}\mathrm{d}t \\ &\leq F(n\pi)+\frac{n+1}{n}\int_0^\pi \frac{\mathrm{d}t}{1+(n\pi)^2\cos^2{t}} \\ &= F(n\pi)+\frac{(n+1)\pi}{n\sqrt{1+n^2\pi^2}}. \end{align} $$ Using the same kind of inequality on the other side it shouldn't be difficult to show that

$$ F(x)\to \frac{1}{2}. $$

Answer 2

For $x=2n\pi$,

$$\int_0^{2n\pi} \dfrac {t}{1+x^2 \cos^2 t}dt=\sum_{k=0}^{n-1}\int_{2k\pi}^{2(k+1)\pi} \dfrac {t}{1+x^2 \cos^2 t}dt=\sum_{k=0}^{n-1}\int_0^{2\pi} \dfrac {t+2k\pi}{1+x^2 \cos^2 t}dt\\ =nI(x)+\pi(n-1)nJ(x)$$ by splitting the integrand.

The integral $I(x)$ can be neglected and we have

$$J(x)=\int_0^{2\pi} \dfrac{dt}{1+x^2 \cos^2 t}=\frac{2\pi}{\sqrt{x^2+1}}=\frac{2\pi}{\sqrt{4n^2\pi^2+1}}\sim\frac1n.$$

The limit should be $\dfrac12$.

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Note that convergence is guaranteed because when $x$ is not a multiple of the period, the extra contribution is neglectible compared to the number of complete periods.