Find $\lim_{k \to \infty}\int_{0}^{\infty}ke^{-kx^2}\arctan(x)dx$ 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?
 A: Enforce the substitution $x=t/\sqrt{k}$.  Then, we can write
$$\int_0^\infty ke^{-kx^2}\arctan(x)\,dx= \int_0^\infty \sqrt{k} e^{-t^2}\arctan(t/\sqrt{k})\,dt$$

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

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!


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 $k\to \infty$. 
