4
$\begingroup$

I was trying to calculate the derivative of the function $$ F(x) =\frac{1}{x}\int_0^x\arctan(e^t)\mathrm{d}t $$ I thought the fastest way was to use the Leibniz's rule for the derivative of a product, $$ (f\cdot g)' = f'g + g'f $$ and, choosing as $f(x) = \frac{1}{x}$ and as $g(x) = \int_0^x\arctan(e^t)\mathrm{d}t$, applying for the second one's derivative the fundamental theorem of calculus, I obtained $$ -\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^t)\right]\Bigg|_{t = 0}^{t = x} = -\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^x)-\frac{\pi}{4}\right] $$ Now there come the problems, since I don't know how to evaluate the limit as $x\rightarrow0$ for the first term of the expression, while the second one, as $x\rightarrow0$, $g'(x)f(x)\rightarrow\frac{1}{2}$. So I plotted the whole thing and I saw something very strange:

[1]: https://i.stack.imgur.com/GE

The blue one is the function (which is right), the red one it's the derivative as calculated before. As you can notice looks like the derivative have a discontinuity in the point 0, while looking at the graph of the function $F(x)$ one would say that there's not such a discontinuity. I tried to evaluate the whole thing with Mathematica, but I did not solve the problem: there are strange things happening at the origin.

Now, there are two possibilities:

  1. The derivative is wrong, but I wonder where, as it's so simple and linear
  2. Grapher app from Mac OS X cannot handle with such functions in a proper way

Can you find out the bug?

$\endgroup$
  • $\begingroup$ Are you sure the blue graph (of $F$) is correct? One of the factors of $F(x)$ is $1/x$ which does have a discontinuity at $x=0$. $\endgroup$ – Zubin Mukerjee Jun 5 '17 at 20:01
  • 2
    $\begingroup$ Shouldn't $d/dx \int_0^x \arctan (e^t) dt = \arctan (e^x)$? $\endgroup$ – sharding4 Jun 5 '17 at 20:03
  • 2
    $\begingroup$ You have a factor of $\pi/4$ that doesn't belong, giving you the singularity. $\endgroup$ – Umberto P. Jun 5 '17 at 20:04
  • 1
    $\begingroup$ you have a mistake in the last part, it should be $0$ instead of $\frac { \pi }{ 4 } $ you forgot derivatives of boundry functions $\endgroup$ – haqnatural Jun 5 '17 at 20:04
  • $\begingroup$ Ok, but apart the factor $\pi/4$ the discontinuity depends on the $-1/x^2 * I$, where $I$ is the integral function $g(x)$... $\endgroup$ – opisthofulax Jun 5 '17 at 20:48
4
$\begingroup$

With your notation $$ g(x)=\int_0^x\arctan(e^t)\,dt $$ we have $$ F'(x)=\frac{xg'(x) - g(x)}{x^2}=\frac{x\arctan(e^x)-g(x)}{x^2} $$ for $x\ne0$. On the other hand, the function $F$ can be extended by continuity at $0$ as $$ \lim_{x\to0}F(x)=\frac{\pi}{4} $$ and $$ \lim_{x\to0}F'(x)= \lim_{x\to0}\frac{1}{2x}\frac{xe^x}{1+e^{2x}}=\frac{1}{4} $$ so $F$ (extended) is also differentiable at $0$.

I used l’Hôpital for both limits.

The bug in your argument is that you wrongly do $$ \frac{d}{dx} g(x) = \Bigl[ \arctan(e^t)\Bigr]_{t=0}^{t=x} $$ instead of $$ \frac{d}{dx}g (x) =\arctan(e^x) $$ according to the fundamental theorem of calculus.

$\endgroup$
  • $\begingroup$ Yes, this was kind of stupid mistake... Anyway beyond is the derivative $-\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^x)\right]$ is right as well as your derivative that you get with the derivative of a function quotient rule $\frac{g(x)}{f(x)}$? Are the solution the same? $\endgroup$ – opisthofulax Jun 5 '17 at 21:06
  • $\begingroup$ @opisthofulax Product rule and quotient rule give the same answer. $\endgroup$ – egreg Jun 5 '17 at 21:07
  • $\begingroup$ Yes I know they give the same result, but our results are different (we have the + and the - inverted, you have $-\frac{1}{x}g(x)$ and $+\frac{1}{x^2}g(x)$ I have the opposite) so I was wondering if someone of we two has make a mistake in the derivative... $\endgroup$ – opisthofulax Jun 5 '17 at 21:22
  • $\begingroup$ ok looks like you've inverted the signs applying the derivative rule for a quotient function, then our two results match perfectly. Thanks for your advices and for your help, sorry it was only a stupid mistake! $\endgroup$ – opisthofulax Jun 5 '17 at 21:31
  • 1
    $\begingroup$ @opisthofulax Everybody makes stupid mistakes, as you saw! ;-) Thanks for the edit. $\endgroup$ – egreg Jun 5 '17 at 21:40
1
$\begingroup$

So the "bug" was in the application of the foundamental theorem of calculus in which does not appear the derivative calculated at the starting point, so that the right derivative is $$ -\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^t)\right]\Bigg|_{t = 0}^{t = x} = -\frac{1}{x^2}\int_0^x\arctan(e^t)\mathrm{d}t + \frac{1}{x}\left[\arctan(e^x)\right] $$

which is the red function in the graph below enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.