About Fundamental Theorem of Calculus Let's say that we have:
$\frac{d}{dx} \int_0^x \log(t) dt$
I want to take the derivative in respect to $x$, as we can see. However wouldn't that be a problem, since we should evaluate $\log$ at $0$, which clearly diverges??
 A: Yes, you're right — this is a bit of a problem. But the issue is not with the Fundamental Theorem of Calculus (FTC), but with that integral. In order to take the derivative of a function (with or without the FTC), we've got to have that function in the first place. So the real question is: do we have a function defined as
$$F(x)=\int_0^x \ln(t)\,dt \quad ?$$
As you correctly noticed, we have a problem here with the lower limit of integration, since $\ln(0)$ is undefined. So this integral does not make sense as a usual integral, but it does as an improper integral — of course, if we also assume that its domain is $x\ge0$.
Let's evaluate this improper integral:
$$F(x)=\int_0^x \ln(t)\,dt=\lim_{b\to0^{+}}\int_b^x \ln(t)\,dt=\lim_{b\to0^{+}}\left.\left(t\ln(t)-t\right)\right|_b^x=\lim_{b\to0^{+}}\left[\left(x\ln(x)-x\right)-\left(\color{red}{b\ln(b)}-b\right)\right]=x\ln(x)-x-\color{red}{0}+0=x\ln(x)-x.$$
The only non-trivial part there was the limit highlighted in red, and it can be shown to be zero using L'Hôpital's Rule:
$$\lim_{b\to0^{+}}b\ln(b)=\lim_{b\to0^{+}}\frac{\ln(b)}{1/b}=\frac{-\infty}{+\infty}=\lim_{b\to0^{+}}\frac{1/b}{-1/b^2}=\lim_{b\to0^{+}}(-b)=0.$$
A: Hint: write the improper integral as the sum of another improper integral plus a definite integral. Apply now the FTC.
