# 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??

• If the (improper) integral is convergent... Dec 23, 2019 at 0:23
• Well, I think that $lim_0 ln(0)*0$ is zero, so it seems not to be a problem. I just wanna be sure...... Dec 23, 2019 at 0:27
• $$F(x)= \int_{0}^{x} \ln (t) dt = [t \ln (t)]_0^x-x$$ which is equal to $$x\ln (x)-x$$ since $lim_{t \to 0} t \ln (t) =0$.
– user726500
Dec 23, 2019 at 0:28
• Even if we don't know an antiderivative, we still can do that -- as long as the integral itself is well-defined. To justify that in case of an improper integral, we need to prove that this improper integral converges. How we do that depends on the actual integral. For example, comparison tests for convergence may help. Dec 23, 2019 at 0:40
• Oh, I have deleted that and asked below your answer hahah sorry. But I got it! Dec 23, 2019 at 0:43

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: $$\lim_{b\to0^{+}}b\ln(b)=\lim_{b\to0^{+}}\frac{\ln(b)}{1/b}=\frac{-\infty}{+\infty}=\lim_{b\to0^{+}}\frac{1/b}{-1/b^2}=\lim_{b\to0^{+}}(-b)=0.$$
• Got it! But how should we proceed if we did not know the antiderivative of the integrand, or at least it's hard to find? Just write: $F(x) = ln(x)1 - lim_0 {ln(0)} 0$ ? Which is the same as $lnx$ Dec 23, 2019 at 0:42