Proving integral $\int_0^1\frac{e^x-1}{x}$ is equal to $\sum_{n=1}^{\infty}\frac{1}{n \cdot n!}$

Question

Show that the following equality is true. $$ \int_0^1\frac{e^x-1}{x}\, \mathrm dx = \sum_{n=1}^{\infty}\frac{1}{n \cdot n!} $$

How can I tackle this problem?

Answer 1

Your integral is not "improper" in the sense that $\frac{e^x-1}{x}\xrightarrow[x\to0]{}1$.

As you noted, for all $x\in\mathbb R$, $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$, so $$\int_0^1 \frac{e^x-1}{x} {\rm d}x = \int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n!}{\rm d}x $$ The convergence under the sum is normal (this you have to prove, and cite the theorem), hence uniform, so you can switch the sum terms : $$\int_0^1 \sum_{n=1}^\infty \frac{x^{n-1}}{n!}{\rm d}x = \sum_{n=1}^\infty \int_0^1 \frac{x^{n-1}}{n!}{\rm d}x = \sum_{n=1}^\infty \left[\frac{x^n}{n\cdot n!}\right]_0^1 = \sum_{n=1}^\infty \frac{1}{n\cdot n!}$$

Answer 2

$$I=\int_0^1\frac{e^x-1}{x}dx$$ Now note that: $$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}=1+x+\frac{x^2}{2}+...$$ So we can say that: $$e^x-1=x+\frac{x^2}{2}+...=\sum_{n=1}^\infty\frac{x^n}{n!}$$ Now we just divide by $x$: $$\frac{e^x-1}{x}=\frac{1}{x}\sum_{n=1}^\infty\frac{x^n}{n!}=\sum_{n=1}^\infty\frac{x^{n-1}}{n!}$$ Which now simplifies our integral to: $$I=\int_0^1\sum_{n=1}^\infty\frac{x^{n-1}}{n!}dx$$ Now interchange the integral and summation: $$I=\sum_{n=1}^\infty\int_0^1\frac{x^{n-1}}{n!}dx=\sum_{n=1}^\infty\left[\frac{x^n}{n.n!}\right]_0^1=\sum_{n=1}^\infty\frac{1}{n.n!}\approx 1.32$$