Proving $\int_0^{+\infty} \frac{\sin(x)}{1+x^4} \ dx$ exists and local Riemann integrable Proving $\displaystyle\int_0^{+\infty} \frac{\sin(x)}{1+x^4} \ dx$ exists
denote $f(x) = \dfrac{\sin(x)}{1+x^4}$, I'm going to assume that $f$ is continuous over the interval $[0,+\infty)$, hence it is locally Riemann integrable,
I have down a proof, but my proof starts off with, since $f$ has no problems at $x = 0 $ we can consider $\displaystyle\int_1^{+\infty} \frac{\sin(x)}{1+x^4} \ dx$ instead - I'm wondering how would I justify this? How could I just prove it is in fact fine at 0.
My other question was what is the difference between locally Riemann integrable and Riemann integrable? I'm staring at the definition but don't get the intuitive difference.
 A: Your $f$ is defined everywhere because $1+x^4$ never vanishes. So $1/(1+x^4)$ is continuous, hence $f$ is continuous everywhere (because $\sin(x)$ is continuous). Thus its integral $\int f$ is locally well defined.
The difference between local and global is the following: consider $f(x)=x$. It is locally integrable but $\int_0^\infty f(x)=\infty$ so it is not "globally" integrable.
If you want indeed to show that $\int f$ exists, you have to check how fast $f(x)$ decreasts as $x\to\infty$ (because, as you said, "at $x=0$ $f$ has no problems")
A: It's fine at zero because the function is continuous on any interval of the form $[0,L]$ for $L>0$. This interval being compact, the function is bounded on any such interval, and therefore is Riemann integrable.
According to the definition I have of 'locally Riemann integrable', this property just guarantees integrability on any compact interval of $\mathbb{R}_{+}$. This is clearly a different notion to integrability on the whole of $\mathbb{R}_{+}$.
To prove that your particular function is locally Riemann integrable, just look at my first comment, and note that any interval $[a,b]$ of $\mathbb{R}_{+}$ is a subinterval of some $[0,L]$. 
To prove that the improper Riemann integral exists, just bound the integrand in absolute value by $\frac{1}{1+x^4}$. This is Lebesgue integrable on $\mathbb{R}$ by the monotone convergence theorem. Let's call your function $f$. Then the integrals of $|f|$ over the sets $[0,L]$ are bounded above uniformly in $L$ by the Lebesgue integral of $|f|$ (which we have just established exists). You then apply the Theorem mentioned here 
http://en.wikipedia.org/wiki/Improper_integral#Improper_Riemann_integrals_and_Lebesgue_integrals
to conclude that the improper integral exists. 
