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May 9th, 2017, 05:19 AM  #1 
Newbie Joined: Dec 2014 From: Italy Posts: 5 Thanks: 0  $a,b\geq 0$ and $a<+\infty$, $b<+\infty$, then $ab<+\infty$?
Let be $f\in C^0(\mathbb{R})$, such that $f\geq 0$. I suppose that $\int_{\mathbb{R}}f(t) dt<+\infty$. Under these hypothesis, can I say that $\int_{0}^{+\infty} f(t) dt\int_{\infty}^{0} f(t)dt<+\infty$ ? I think yes, since $\int_{\mathbb{R}}f(t) dt<+\infty$ iff $\int_{0}^{\infty}f(t)dt<+\infty$ e $\int_{\infty}^{0}f(t)dt<+\infty$. If $a,b\geq 0$ and $a<+\infty$, $b<+\infty$, then $ab<+\infty$. The viceversa of this last statement is not true, right? Am I right? Thanks! Last edited by Glo; May 9th, 2017 at 05:47 AM. Reason: more explicative title 
May 19th, 2017, 03:38 AM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,576 Thanks: 668 
Specifically, and we are told that the right side is finite.

May 19th, 2017, 06:10 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra 
Suppose that $\displaystyle \int_0^{+\infty} f(t)\,\mathrm dt = 1$ and $\displaystyle \int_{\infty}^0 f(t)\,\mathrm dt = 1$ then $$\begin{align*} \left \int_0^{+\infty} f(t)\,\mathrm dt \int_{\infty}^0 f(t)\,\mathrm dt \right &= 1 + 1 = 2 \\ \left \int_{\infty}^{+\infty} f(t)\,\mathrm dt \right &= 0 \end{align*}$$


Tags 
functions, integrability, integrable, sum 
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