Consider the following integral

$\int_{0}^{1} f(x) dx$

If $f(x)= \text{exp}(x)/ \sqrt{x}$, we have an integrable singularity at the origin.

If $f(x)= \text{exp}(x)/ x$, we have a non-integrable singularity at the origin.

There is no point in attempting to devise a strategy for estimating $\int_{0}^{1} \text{exp}(x) dx/x$, because the integral does not exist.

I am a layman for numerical analysis. I do not understand why the integrand is non-integrable if $f(x)= \text{exp}(x)/ x$ and why there is no point for estimating $\int_{0}^{1} \text{exp}(x) dx/x$?

  • $\begingroup$ What do you get, if you write $\exp(x)=1+x + x^2/2 \cdots$ and integrate termwise? What is the antiderivative of $1/x$? $\endgroup$ – gammatester Sep 15 '17 at 13:40

It is not a question of numerical analysis, it is a question of calculus. Near the origin $\exp(x)\approx 1$, so we can just talk about $\int_0^a \frac 1x\;dx$ and $\int_0^a \frac 1{\sqrt x}\; dx$ for $a$ rather small. The $\exp(x)$ factor will change the value but not whether the integral converges. Because the integrand diverges at zero, for the first we are really considering $\lim_{b\to 0}\int_b^a \frac 1x \; dx=\lim_{b \to 0}\log a - \log b$, wbich diverges. For the second, we are considering $\lim_{b\to 0}\int_b^a \frac 1{\sqrt x} \; dx=\lim_{b \to 0}2(\sqrt a -\sqrt b)=2\sqrt a$ and is nicely convergent.


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