I don't understand what should I do in this problem: "Using the Schwartz inequality in $L^2(0, \infty)$ find a increment (increase) of the modulus of: $$\int_{0}^\infty \frac{x^{3/2}}{1+ix}e^{-x} dx$$ "
I suppose that this is a product in $L^2(0,\infty):$ $$\langle f,g\rangle=\int_{0}^\infty \overline{f}g \, dx.$$ But who's $f$, who's $g$? I've tried various combinations but for none of them can I get, for both $f$ and $g$: $$\int_{0}^\infty |f|^2dx<\infty.$$ I want to show that Schwartz the inequality holds: $$|\langle f,g\rangle| = \left|\int_{0}^\infty \frac{x^{3/2}}{1+ix}e^{-x}\, dx \right|\le ||f||\cdot||g||=\sqrt{\int_{0}^\infty |f|^2 \,dx}\cdot\sqrt{\int_{0}^\infty |g|^2 \, dx}.$$ where it is essential for the functions to be part of $L^2$. Where am I wrong?
You need to group your integrand into two functions, each of which is $L^2(0,\infty)$. Some quick calculations yield that $f(x)=\dfrac{1}{1+ix}\in L^2(0,\infty)$, and it's clear that $g(x)=x^{3/2}\,e^{-x}\in L^2(0,\infty)$. We have that $\|f\|=\dfrac{\pi}{2}$, and $\|g\|=\dfrac38$. Can you finish from here?