Show $\phi_j(x)=\frac{\phi_{j-1}(x)-\phi_{j-1}(0)}{x}$ for $\phi_j(x):=\frac{1}{(j-1)!}\int_{0}^{1}e^{(1-\theta)x}\theta^{j-1}d\theta$

Question

Consider

$$\phi_j(x):=\frac{1}{(j-1)!}\int_{0}^{1}e^{(1-\theta)x}\theta^{j-1}d\theta, \ \ \ \ \ j \ge 1, \ x \in \mathbb R \tag 1$$

$$\phi_0(x):=e^x$$

(which is used in numerical procedures involving differential equations)

Show the recursion formula

$$\phi_j(x)=\frac{\phi_{j-1}(x)-\phi_{j-1}(0)}{x}, \ \ \ \ \ j \ge 1, \ \ x \neq 0$$

What are the values of $\phi_{j-1}(0)$

I tried to show it by induction.

$j=1$:

$\phi_1(x)=\frac{\phi_0(x)-\phi_0(0)}{x}=\frac{e^x-1}{x}$

And $\int_{0}^{1}e^{(1-\theta)x}d\theta=\frac{e^x-1}{x}$

But at the induction step $j \to j+1$ I run into problems

$\phi_{j+1}(x)=\frac{\phi_{j}(x)-\phi_{j}(0)}{x}$. Then, by induction hypothesis I plugged in $(1)$ for $\phi_j(x)$ and $\phi_j(0)$ but this only ended in a complicated integral that my CAS could'nt solve.

If there is a other way than induction, that's also fine.

Answer 1

Let $j \geq 1$, then, integrating by part (the bracket term is zero at $0$ because of $t^j$, at $1$ because of the $e^{(1-t)x}-1$ term), $\frac{(j-1)!}{x}(\phi_j(x)-\phi_j(0))=\int_0^1{t^{j-1}\frac{(e^{(1-t)x}-1}{x}\,dt}=-\frac{1}{j}\int_0^1{t^j(-xe^{(1-t)x}\,dt}=(j-1)!\phi_{j+1}(x)$.