Explicit formula for recursive formula of an integral

Question

I've been trying to find an explicit formula for the integral $\displaystyle I_n= \frac{1}{e}\int_0^1 e^xx^{n-1}\, dx = 1-(n-1)I_{n-1},n=2,3,4,..., I_1=1-\frac{1}{e}$

Firstly, I tried to construct a $2\times2$ matrix and see if it can be diagonalized, but it seems we cannot do that as $I_n$ is dependend only on the previous term and not the two previous terms.

Then, I tried to see if there can be an explicit formula easily identified laying out the first terms as such: \begin{array}\\I_n:&&1-\frac{1}{e},&&\frac{1}{e},&&1-\frac{2}{e},&&-2+\frac{6}{e},&&9-\frac{24}{e},&&-44+\frac{95}{e} \end{array}

But I can't seem to find any correlation between $n$ and $I_n$.

Thanks in advance.

Answer 1

Recall the following:

$$\int_0^1 e^{xt}\ dx=\frac{e^t-1}{t}$$

It thus follows that

$$\int_0^1xe^{xt}\ dx=\frac d{dt}\frac{e^t-1}{t}$$

$$\int_0^1x^2e^{xt}\ dx=\frac{d^2}{dt^2}\frac{e^t-1}{t}$$

$$\int_0^1x^ne^{xt}\ dx=\frac{d^n}{dt^n}\frac{e^t-1}{t}$$

By applying Leibniz's rule for nth derivative of a product,

$$\frac{d^n}{dt^n}\frac{e^t}{t}-\frac1t=\frac{(-1)^{n+1}n!}{t^{n+1}}+e^t\sum_{k=0}^n\binom nk\frac{(-1)^kk!}{t^{k+1}}$$

And at $t=1$, we get

$$\int_0^1x^ne^x\ dx=(-1)^{n+1}n!+e\sum_{k=0}^n\binom nk(-1)^kk!$$

Answer 2

HINT: using integration by parts we have that

$$I_n=\int x^n e^x\mathrm dx=x^ne^x-n\int x^{n-1}e^x\mathrm dx=\ldots=e^x\sum_{k=0}^n(-1)^{k}x^{n-k}n^\underline k$$

where $n^\underline k$ is a falling factorial. Then we have the recursion

$$I_n=x^ne^x-nI_{n-1}$$

P.S.: the above summation seems not closable in it general setup (checking it in mathematica it have the form of an incomplete gamma function when $x>0$) but probably you can get a closed expression for some definite integral.