Find $\lim_{k \to \infty}\int_{0}^{\infty}ke^{-kx^2}\arctan(x)dx$

Question

Find $\lim_{k \to \infty}\int_{0}^{\infty}ke^{-kx^2}\arctan(x)dx$.

I think that the limit is infinity.

$ke^{-kx^2}\leq ke^{-kx^2}\arctan(x)$ for $[\tan(1),\infty)$, but by integrating we know that $\int_{0}^{\infty}ke^{-kx^2}\to \infty$ and so out sequence of original integrals diveres too.

Is this correct?

Answer 1

Enforce the substitution $x=t/\sqrt{k}$. Then, we can write $$\int_0^\infty ke^{-kx^2}\arctan(x)\,dx= \int_0^\infty \sqrt{k} e^{-t^2}\arctan(t/\sqrt{k})\,dt$$

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Now, $\arctan(x)\le x$ for $x\ge 0$, so that $\left|\sqrt{k} e^{-t^2}\arctan(t/\sqrt{k})\right|\le te^{-t^2}$.

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Inasmuch as $\int_0^\infty te^{-t^2}\,dt<\infty$, the Dominated Convergence Theorem guarantees that

$$\begin{align} \lim_{k \to \infty}\int_0^\infty \sqrt{k} e^{-t^2}\arctan(t/\sqrt{k})\,dt&=\int_0^\infty \lim_{k\to\infty}\left(\sqrt{k} e^{-t^2}\arctan(t/\sqrt{k})\right)\,dt\\\\ &=\int_0^\infty te^{-t^2}\,dt\\\\ &=\frac12 \end{align}$$ And we are done!

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NOTE: Analyzing the argument in the OP

The OP observed correctly that for $x\in [\tan(1),\infty)$, $\arctan(x)>1$.

However, the integral $\displaystyle \lim_{k\to \infty}\int_{\tan(1)}^\infty ke^{-kx^2}\,dx\ne \infty$.

To see this, simply note that $ke^{-kx^2}\le \frac{1}{x^2}$ and $\int_{\tan(1)}^\infty \frac{1}{x^2}\,dx=\cot(1)$.

However, the $\lim_{k\to\infty}\int_0^{\tan(1)}ke^{-kx^2}\,dx=\infty$.

But for $x\in [0,\tan(1)]$, $\arctan(x)$ does not behave like a constant. Rather, it has a small argument approximation $\arctan(x)\sim x$. And clearly $\int_0^{\tan(1) }kx e^{-kx^2}\,dx$ converges as $k\to \infty$.