I computed the integral of $\sqrt{1+e^x}$ by hand and got $$2\sqrt{1+e^x} + \ln\left(\sqrt{1+e^x} - 1\right) - \ln\left(\sqrt{1+e^x} + 1\right) + C,$$ or $$2\sqrt{1+e^x} + \ln\left(\frac{\left(\sqrt{1+e^x} - 1\right)^2}{e^x}\right) + C.$$ But when I plot the graph on Desmos, this strange thing happened:

Desmos Graph of the function

Can anyone explain this discontinuity phenomenon I have seen? And, is there a better form of the function that the one I have?

  • 1
    $\begingroup$ For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$
    – Martin R
    Jun 16, 2022 at 8:32
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    $\begingroup$ Looks like a problem due to limited accuracy of used arithmetic. IEEE-754 double? Might be due to bad numerical conditioning that precision of arithmetic is exhausted. $\endgroup$ Jun 16, 2022 at 8:49
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    $\begingroup$ as @emacsdrivesmenuts said, there is no reason why this function should be discontinuous within your domain. It is likely due to the fact that you are dealing with the logarithm of very small numbers and the way desmos calculates this $\endgroup$
    – Henry Lee
    Jun 16, 2022 at 9:58

1 Answer 1


The problem is very likely due to the finite precision of the arithmetic that's used (presumably IEEE-754 double), together with bad conditioning due to cancellation:

In $\ln(\sqrt{1+e^x}-1)$, when $x$ is small ($x\to-\infty$) then $e^x\to 0$ and the square-root evaluates to $\approx 1+e^x/2$. The magnitude of that term is $\approx 1$, and when you subtract $1$ then what you get is $e^x/2$ but only with the precision in which $1$ was represented. This is because the precision of $1+e^x/2$ is basically the precision of $1$ when $e^x\to0$.

When you coalesced the two $\ln$ terms, you got

$$\begin{align} \ell(x) = \ln\left(\sqrt{1+e^x} - 1\right) - \ln\left(\sqrt{1+e^x} + 1\right) &= \ln\left(\frac{\sqrt{1+e^x} - 1}{\sqrt{1+e^x} + 1}\right) \\ &= \ln\left(\frac{(\sqrt{1+e^x} - 1)^2}{e^x}\right) \end{align}$$ which made the problem of cancellation in $\sqrt{1+e^x}-1$ even worse because you squared that term to get a "nice" denominator. Provided $e^x$ is small, squaring that term means doubling $x$, thus $x$ gets even more negative which amplifies the problem.

is there a better form of the function that the one I have?

Going for a nice numerator, however, gives:

$$\begin{align} \ell(x) &= \ln\left(\frac{e^x}{(\sqrt{1+e^x} + 1)^2}\right) \\ &= x - 2\ln(\sqrt{1+e^x}+1) \\\tag 2 \end{align}$$ The representation in $(2)$ mitigated the problem of cancellation, and it's still a nice one. The nasty $\sqrt{1+e^x}-1$ has disappeared altogether.

Note: I cross-checked my result against Desmos. The artifact is around $x=-35$:


Image showing breakdown of accuracy


Where to expect the breakdown

The smallest value greater than $1$ that can be represented by IEEE-754 double is $1+\varepsilon$ with $\varepsilon=2^{-52}$. Determining $x$ such that $\sqrt{1+e^x}-1 = \varepsilon$ means $$1+e^x = (1+\varepsilon)^2 \approx 1+2\varepsilon$$ $$x \approx \ln (2\varepsilon) \approx -51\ln2\approx-35.4$$

And you actually observed a breakdown at $x\approx-35$.

  • $\begingroup$ Thank you so much! $\endgroup$ Jun 17, 2022 at 11:52

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