Finding integrating the sequence of functions $f_n(x)=\frac{2nx}{1+n^2x^4}$

Question

If $f(x)=\lim_{n \to \infty}f_n(x)$ where $f_n(x)=\dfrac{2nx}{1+n^2x^4}$, find $$\int_0^1f(x)dx\quad \text{and} \quad \lim_{n \to \infty}\int_0^1f_n(x)\,dx$$

For the first part:

$$\lim_{n\to \infty}f_n(x)=0$$

Therefore

$$\int_0^1f(x)\,dx=\int_0^10\,dx=0$$

Then for the second part:

$$\lim_{n\to \infty}\int_0^1f_n(x)\,dx=\lim_{n\to \infty}2n\int_0^1\frac{x}{1+n^2x^4}\,dx=\lim_{n\to \infty}\left(\tan^{-1}(n)-\tan^{-1}(0)\right)=?$$

It looks like it goes to $\pi/2$ since $\tan^{-1}$ is an integer when we're at $\pi/4,5\pi/4,9\pi/4,\ldots$

and these are a distance of $\pi/4$ from $\tan^{-1}(0)$. Then we get $\pi/2$. But I dont see how else to show this or if my calculations are correct. Does this make sense? Is there a better way to show this?

Answer 1

$$\int\limits_0^1\frac{2nx}{1+n^2x^4}dx=\int\limits_0^1\frac{d(nx^2)}{1+(nx^2)^2}=\left.\arctan nx^2\right|_0^1=\left(\arctan n-\arctan 0\right)=$$

$$=\arctan n\xrightarrow[n\to\infty]{}\frac{\pi}{2}$$

Answer 2

Use the substitution $x = y/\sqrt{n}$ to get

$$ \begin{align} 2 \int_0^1 \frac{nx}{1+n^2 x^4}\,dx &= 2 \int_0^\sqrt{n} \frac{y}{1+y^4}\,dy \\ &\to 2 \int_0^\infty \frac{y}{1+y^4}\,dy \\ &= \frac{\pi}{2}. \end{align} $$

(The last integral can be found here.)