The exponential integral is defined as: $$E_n=x^{n-1} \int_x^\infty{\frac{e^{-t}}{t^n}}dt.$$
I want to prove the relation: $$\int_0^xE_1(t)dt=1-E_2(x)$$
Integrating $E_{n-1}=-\frac{dE_n}{dx}$ from $0$ to $x$ with $n=2$, I got to: $$\int_0^xE_1(t)dt=E_2(0)-E_2(x).$$
I can't see how $E_2(0)$ will equal $1$. Looking at the definition, it should equal $0$.
Can you please point out what I did wrong?
Best Answer
We have: $$\lim_{\tau \to 0}E_2(\tau)=\lim_{\tau \to 0}\bigl(\tau \int_\tau^\infty{\frac{e^{-t}}{t^2}}dt\bigr).$$
Now let $t= \tau u$ and change the limits of integration: $$\lim_{\tau \to 0}E_2(\tau)=\lim_{\tau \to 0}\bigl(\tau \int_{1}^\infty{\frac{e^{-\tau u}}{\tau^2u^2}}\tau du\bigr)= \int_{1}^\infty{\frac{du}{u^2}},$$ which is easily seen to be $\color{red}{1}$.