Evaluate the integral $\int e^{e^x + 2014x}dx$

Question

Evaluate the integral $\int e^{e^x + 2014x}dx$. If you try to find first derivative of $e^{e^x}$ you'll get $e^{e^x}e^{x}$. The second is $e^{e^x}e^{2x} + e^{e^x}e^{x}$ and so on. For the $2014$-th derivative you'll get $e^{e^x + 2014x}$ as one of it's members. If you try to make a substitution you'll get $\int e^tt^{2013}dt$ which still looks bad even knowing formula $\int e^{ax}x^ndx = \frac{x^ne^x}{a} + \frac{n}{a}\int x^{n-1}e^{ax}dx$

Answer 1

The substitution is the way to go. It doesn't look too terrible if you write it abstractly. If $p$ is a polynomial, then $(e^tp(t))' = e^t(p(t) + p'(t))$, so to get a primitive of $e^t t^n$, you need a polynomial $p_n$ with $p_n(t) + p_n'(t) = t^n$. Computing the first few explicitly leads to the formula

$$p_n(t) = \sum_{k=0}^n (-1)^k\frac{n!}{(n-k)!}t^{n-k} = \sum_{k=0}^n (-1)^{n-k}\frac{n!}{k!}t^k.$$

Setting $n = 2013$ and substituting back, we find

$$\int e^{e^x+2014x}\,dx = e^{e^x}\sum_{k=0}^{2013} (-1)^{k+1}\frac{2013!}{k!}e^{kx} + C.$$

Answer 2

Your approach using $t=e^x$ should be useful. As you noted, by the reduction formula you posted:

$$\begin{align} \\ \int t^{2013} e^{t}dt &= t^{2013}e^t - 2013 \int t^{2012} e^t \\ &= t^{2013}e^t - 2013 \left(t^{2012}e^t - 2012\int t^{2011} e^t \right) \\ &= t^{2013}e^t - 2013 \left(t^{2012}e^t - 2012 \left(t^{2011}e^t - 2011 \int t^{2010} e^t \right)\right) \\ &= t^{2013}e^t - 2013 t^{2012}e^t + 2013\cdot 2012 t^{2011}e^t - 2013\cdot 2012\cdot2011 \int t^{2010} e^t \end{align}$$

You should be able to start seeing a pattern here. If you continue expanding, you will have:

$$\begin{align}\\ &= t^{2013}e^t - 2013 t^{2012}e^t + 2013\cdot 2012 t^{2011}e^t - \cdots - \frac{2013!}{2!} t^2 e^t + \frac{2013!}{1!} t e^t - 2013! e^t + C\\ &= \sum_{i=0}^{2013} (-1)^{i+1} \frac{2013!}{i!}t^ie^t + C\\ &= 2013! e^t \sum_{i=0}^{2013} (-1)^{i+1} \frac{t^i}{i!} + C \end{align}$$

Substituting back in, you have:

$$2013! e^{e^x} \sum_{i=0}^{2013} (-1)^{i+1} \frac{e^{ix}}{i!} + C$$

To be a bit more rigorous, you should be able to prove by induction that for any positive integer $n$:

$$\int e^{e^x + nx}dt = (n-1)! e^{e^x} \sum_{i=0}^{(n-1)!} (-1)^{i+1} \frac{e^{ix}}{i!} + C$$