Weird integrals calculation

Question

How to calculate these weird integrals (if it is possible): $$\int e^{dx}$$ and $$\int x^{dx}?$$ Thanks

Answer 1

We might extend the definition of an integral by saying that $$\int_a^b f(x)g(dx)=\lim_{n\to\infty}\sum_{i=0}^{n-1}f\bigg(a+\frac{b-a}{n}i\bigg)\cdot g\bigg(\frac{1}{n}\bigg)$$ However, if you let $g(x)=e^x$, this sum always diverges because $g(1/n)$ does not approach zero as $n\to\infty$. Thus, the integral $$\int_a^b e^{dx}$$ would diverge if defined in the above sense. However, the integral $$\int_a^b (e^{dx}-1)=^{?}\space 1$$ could be evaluated using the tentative definition above by noticing that $e^{1/n}-1\sim \frac{1}{n}+\frac{1}{2n^2}+...$.

For something similar, see my answer to this previous question for an explanation of why integrals like $$\int_a^b (dx)^2$$ and $$\int_a^b (dx)^n,\space\space n>1$$ are equal to zero for all values of $a,b$.

Answer 2

Comparing Taylor series with $df=f^\prime dx$ gvies $dx^2=0$. Thus $\int e^{dx}=\int(1+dx)$ makes no sense; nor does $\int x^{dx}=\int\exp(dx\ln x)=\int(1+dx\ln x)$. But$$\int\left(e^{dx}-1\right)=\int dx=x+C,\,\int(x^{dx}-1)=x\ln x-x+C.$$