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 $\frac{1}{16}$?

  • $\begingroup$ 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 $0$. $\endgroup$
    – Paramanand Singh
    Apr 30, 2021 at 2:52
  • $\begingroup$ 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? $\endgroup$
    – pawel
    Apr 30, 2021 at 6:43
  • 1
    $\begingroup$ 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 $L$. $\endgroup$
    – Paramanand Singh
    Apr 30, 2021 at 8:20

1 Answer 1


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 $x = 1$.)


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .