prove the equivalence of integral asymptotic $\int_0^{\frac{1}{a}}e^x\arctan(a^3x)dx = \frac{\pi}{2a}+O(\frac{1}{a^2})$, where $a\rightarrow +\infty$
I've tried to integrate by parts and make a change $e^x = 1+x+O(x^2)$ and $arctg(x) = \frac{\pi}{2}-\frac{1}{x}+O(\frac{1}{x^3})$, but didn't succeed. Can you give me a hint, please?
 A: First, let us replace $a = \frac{1}{b}$ (and $b \to 0$). Then we can split integral in two:
$\int_0^{b^2} e^x \arctan(\frac{x}{b^3})\, dx + \int_{b^2}^b e^x \arctan(\frac{x}{b^3})\, dx$. The function in the first integral is bounded by say $\frac{e \pi}{2}$, so the first integral value is $O(b^2)$.
For the second integral, we can write$\int_{b^2}^b e^x \arctan(\frac{x}{b^3})\, dx = \int_{b^2}^b e^x \frac{\pi}{2}\, dx + \int_{b^2}^b e^x \left(\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right)\, dx$. The former is $\frac{\pi b}{2}$.
Now we can use your asymptotic (better with explicit constant): $\left|\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right| < \frac{c b^3}{x}$ for some constant $c$.
$\int_{b^2}^b e^x \left(\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right)\, dx \leqslant
\int_{b^2}^b e^x \left|\arctan(\frac{x}{b^3}) - \frac{\pi}{2}\right|\, dx \leqslant
\int_{b^2}^b \frac{cb^3 e^x}{x}\, dx$. This can be computed explicitly, or can be upper bounded by $c b^3 \ln b = O(b^2)$ by noting $e^x$ is bounded by $1$.
