# Continuity of a piecewise integral function

I have a problem in evaluating the limit as $$x$$ goes to $$0$$ of: $$F(x)=\int_1^x f(t)\, dt\,\,\,\ \text{with }f(t)\begin{cases}t^3\ln{t}\, \text{ for }t>0\\ \arctan{t} \, \text{ for } t\leq 0\end{cases}$$ Now I have remarked that since $$\lim_{t\to 0^+}f(t)=0=f(0)$$ then the function $$f$$ can be continously extended in $$0$$ and so the integral function will be continous and defined in $$\mathbb{R}$$.

If it is continous in $$0$$ this means that $$F(x)\to F(0)=\int_1^0t^3\ln{t}\,dx=-\int_0^1t^3\ln{t}\,dx=\frac{1}{16}$$.

$$\textbf{Problem}$$ $$\lim_{x\to 0^-}\int_{1}^x f(t)=-\lim_{x\to 0^-}\int_{x}^1 f(t)=???$$ In fact $$x\to 0^-$$ so I have that the extreme $$x$$ tends to $$0^-<0$$ but the second is $$1$$...how can I write this limit and check that it is equal to $$\frac{1}{16}$$?

• Whatever be the integrand $f(t)$ the first thing to note is that the integral function $F(x) =\int_1^x f(t) \, dt$ is always continuous. So there is no need to check for continuity of integrand at $0$. Apr 30, 2021 at 2:52
• Ah ok... but the fact that $f$ is continous in $0$ is useful for me to state that $F'(x)=f(x)$ $\forall x\in\mathbb{R}$, right? Apr 30, 2021 at 6:43
• Yeah, more generally if $f(x) \to L$ as $x\to c$ then $F'(c) =L$. And it happens for one sided limits as well. If right (left) hand limit of $f$ at $c$ is $L$ then right (left) hand derivative of $F$ at $c$ is $L$. Apr 30, 2021 at 8:20

If $$x < 0$$, you have \begin{align*} F(x) &= \int_{1}^{x} f(t)\, dt \\ &= \int_{1}^{0} f(t)\, dt + \int_{0}^{x} f(t)\, dt \\ &= \frac{1}{16} + \int_{0}^{x} \arctan t\, dt. \end{align*}
First problem: Because $$F$$ is continuous (as are all definite integrals), you know $$\lim(F, 0^{-}) = \lim(F, 0^{+}) = \frac{1}{16}$$. If you like, this gives you the constant of integration when you antidifferentiate $$\arctan$$.
Second problem: Your explicit form is not continuous at $$0$$; the constant in the numerator of the $$x \leq 0$$ case must agree with $$F(0) = \frac{1}{16}$$. (And if it matters, you don't need a separate case for $$x = 1$$.)