Is $\int z^n e^{az}dz $ a combination of exponentials and polynomials?

Question

We have $$I(n)=\int z^n e^{az}dz=\int z^n \left (\frac{1}{a}e^{az}\right )'dz=\frac{1}{a}z^ne^{az}-\frac{1}{a}\int nz^{n-1}e^{az}dz \\ \Rightarrow I(n)=\frac{1}{a}z^ne^{az}-\frac{1}{a}nI(n-1) \ \ \text{ with } \ \ I(1)=\int ze^{az}dz=\frac{1}{a}ze^{az}-\frac{1}{a^2}e^{az}$$

When we solve this recursive relation, we get an element of the ring $\mathbb{C}[z, e^{\lambda z}\mid \lambda \in \mathbb{C}]$, so an expression that contains exponentials, polynomials and their products.

Is this correct?

Answer 1

From your recursive relation, $$ I_n=\frac{1}{a}z^ne^{az}-\frac{1}{a}nI_{n-1} $$ multiplying by $(-1)^na^n\dfrac1{n!} $ you easily get $$ (-1)^na^n\frac{I_n}{n!}=(-1)^na^{n-1}\frac{z^n}{n!}e^{az}+(-1)^{n-1}a^{n-1}\frac{I_{n-1}}{(n-1)!} $$ or $$ (-1)^na^n\frac{I_n}{n!}-(-1)^{n-1}a^{n-1}\frac{I_{n-1}}{(n-1)!}=(-1)^na^{n-1}\frac{z^n}{n!}e^{az} $$ then by summing, terms telescope and we get $$ (-1)^na^n\frac{I_n}{n!}-I_0=\sum_{k=1}^n(-1)^ka^{k-1}\frac{z^k}{k!}e^{az} $$ that is

$$ (-1)^na^n\frac{I_n}{n!}=e^{az} \sum_{k=0}^n(-1)^ka^{k-1}\frac{z^k}{k!}\in \mathbb{C}[z, e^{a z}] $$

as announced.

Answer 2

Yes, and in fact you can write down a generating function describing which combination. Write

$$I_n = \int_0^z x^n e^{ax} \, dx.$$

Then

$$\sum_{n \ge 0} I_n \frac{t^n}{n!} = \int_0^z e^{(a+t)x} \, dx = \frac{1}{a + t} \left( e^{(a+t)z} - 1 \right).$$

You can recover Olivier's exact formula by expanding this out a bit as a power series in $t$.