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Why is the integrand bounded? The only way I can see that it is bounded is if $x$ never is equal to 0, but if this is true (that $x\neq 0$) then where is this indicated that we don't use the endpoint?

"Open the question so I can answer it as I know the answer to it now"- also I am not sure what to do it open it

Reuben
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  • $\int_0^{c} \frac{\sin(t)}{t}dt$ is in fact the integral of the function $f$ defined by $f(0)=0$ (or any other value) and for $t\neq 0$, $f(t)= \frac{\sin(t)}{t}$. – Jean-Claude Colette Aug 03 '23 at 19:05
  • For some integrals, an extension by continuity of the function is interesting, because it allows to justify the convergence. For $t\mapsto \frac{\sin t}{t}$, we have to define $f(0):=1$. – Jean-Claude Colette Aug 03 '23 at 19:11
  • $\lim_{x\to 0}\frac{\sin \lambda x}x = \lambda$ so the integrand is bounded above by $\lambda(f(x) - f(0+))$ which goes to $0$ at $x \to 0+$. This part you were intended to unsderstand without comment. Their comments address why it remains bounded as $\lambda \to \infty$. You should also be aware that even the Riemann integral is not affected by a function taking on a different value at a single point. – Paul Sinclair Aug 05 '23 at 02:13
  • @PaulSinclair, thanks for the reply. I belive we cant assume that $f(x)-f(0+)$ approaches 0 faster than $\lambda$ to infinity, which in turn will make the integrand infinite. I then supposed that by moving the $x$ under $f(x)-f(0+)$ instead of $sin$ like they did we can create "sort of" a one sided derivative by which stated in the theorem that $f$ is piecewise smooth so we know now that $\frac{f(x)-f(0+)}{x}$ is bounded as $x$ tends to 0. The main problem comes from that Riemann integral can use the limits of an integral and $\frac{f(0)-f(0+)}{0}$ is not defined? – Reuben Aug 06 '23 at 23:41
  • correction, not one sided derivative but a derivative about a point $p \rightarrow 0^+$. – Reuben Aug 07 '23 at 00:33
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    There is no race to see which goes where faster. They are arguing that the integrand has a bound $M$ that is independent of $\lambda$, and thus the integral is bounded by $Mb$ for all $\lambda$, So, no, it cannot become infinite. Note that $|\sin \lambda x| \le 1$ for all $x$ and $\lambda$. And once again, please remember that even the Riemann integral is not affected by changes at a single point. So no, it does not matter if the integrand is undefined at $0$, as long as it is well-behaves everywhere else. – Paul Sinclair Aug 07 '23 at 05:20
  • One other point: my comments are a bit handwavy because I do not know the exact definition the author uses for "piecewise smooth". For the most part this is obvious, but not entirely at the breaks. For example, would $f(x) = \begin{cases}1/x&x>0\0&x\le 0\end{cases}$ count as piecewise smooth? By the assumption that $f(0+)$ exists in the argument, I assume not. Also I do not know what Lemma 6.10.1 is, particularly what conditions it requires. So I am forced to guess at the author's argument. – Paul Sinclair Aug 07 '23 at 15:58

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