# Riemann Integrable Function Problem

I have recently read a book on real analysis, more specific the part about Riemann Integrals. It says that if a function is bounded in a interval $$[a,b]$$, then it is integrable and also when a function has discontinuities such their measure is zero, then it is integrable. But the following examples are driving me crazy:

$$\lim_{t \to 0}\int_t^x \ln(s)ds\,\,\,, \,\,\, \lim_{t \to 0}\int_t^x \frac{ds}{s}$$

The left one is integrable because the discontinuity at zero has measure zero. Fine. But what about the right one? What happens? LET'S SUPPOSE WE DO NOT KNOW THEIR ANTIDERIVATIVES!

Thanks

• Neither function is bounded on $[0,x]$ (nor are they even defined at $0$... but anyway, there's no way to extend it to a bounded function on $[0,x]$). However, for every $\alpha>0$ and $\beta > \alpha$, then the two functions you mention are continuous and bounded on $[\alpha, \beta]$, hence Riemann integrable on $[\alpha, \beta]$. – peek-a-boo May 29 at 2:19
• I see, but why the left converges and the right does not. As is in the question, let's suppose we do not know their anti-derivatives. How could we be sure that they are integrable or not – Mr. N May 29 at 2:59

Neither function is Riemann integrable. $$\int_0^xln(s)ds=\lim_{t\to 0}\int_t^xln(s)ds=xln(x)-x$$, where the latter is Riemann integrable.
$$\int_0^x\frac{1}{s}ds=\lim_{t\to 0}\int_t^x\frac{1}{s}ds=\lim_{t\to 0}(ln(x)-ln(t))$$ which is infinite.