Studying the derivative of the integral function $\frac{1}{x}\int_0^x\arctan(e^t)\mathrm{d}t$

Question

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:

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?

Answer 1

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.

Answer 2

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