# Schwartz inequality for $|\int_{0}^\infty \frac{x^{3/2}}{1+ix}e^{-x} dx|$

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?

• Your question is confusing: are you trying to prove the Schwartz inequality, or are you trying to use it to find an increment of the integral you mentioned? Also, what is an increment of an integral? Is that towards differentiating the integral? – Adrian Keister Feb 19 '18 at 18:40
• @AdrianKeister I know it's confusing, I don't understand what should I do. In italian we say "majoration", i'll give you an example to let you understand what I mean since I can't think to an english term: $|sinx|<1$. The integral I think that could be the inner product in $L^2$ – Gitana Feb 19 '18 at 20:41
• Ok, I think I know what you're trying to do. In English, we would say that you're trying to "bound" the absolute value of the integral. I've suggested some edits that should help. – Adrian Keister Feb 19 '18 at 20:50

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?
• I have one only question: shouldn't be $f=\frac{1+ix}{1+x}$? – Gitana Feb 19 '18 at 21:03
• since $(f,g)=\int_0^\infty \overline{f}g dx$ – Gitana Feb 19 '18 at 21:04
• Nope! We have to carve out two functions, $f$ and $g$, from your integrand, that, when multiplied together, reproduce your integrand. Both of those functions need to be $L^2$. – Adrian Keister Feb 19 '18 at 21:05
• Note that when we calculate the norm of $f$, we do $\|f\|=\displaystyle\int_0^{\infty}\overline{f} \, f \, dx,$ and that's where we need the complex conjugate. – Adrian Keister Feb 19 '18 at 21:05