Integrating x^ne^xn: Analyzing & Solving

[Mayhem]

I was bored and tried to integrate $x^n e^{xn}$. I seem to be on the right track, but ultimately it is not entirely correct. Here is my work:

Given is the integral
$$I = \int x^ne^{nx}dx$$

where $n \geq 1$

We substitute $t = nx$ which gives us $\frac{dt}{dx} = n \Rightarrow dx = \frac{1}{n}dt$ and $x = \frac{t}{n}$

Plugging that in, we obtain
$$ I = \int \left(\frac{t}{n}\right)^ne^t \frac{1}{n}dt $$
which can be reduced to
$$ I = \frac{1}{n^{n+1}} \int t^ne^t dt $$
The expression inside of the integral can be integrated by parts if we simply integrate $n$ times, letting $dv = e^t$ every single time, the degree of the t-polynomial will decrease to zero and cancel out to 1.

We may do this for the first few steps to obtain a pattern:
$$I =\frac{1}{n^{n+1}} \left( e^tt^n - \int nt^{n-1}e^t dt \right ) $$
$$=\frac{1}{n^{n+1}} \left( e^tt^n - \left (nt^{n-1}e^t - \int n(n-1)t^{n-2}e^t dt \right ) \right) $$
$$=\frac{1}{n^{n+1}} \left( e^tt^n - \left (nt^{n-1}e^t - \left ( n(n-1)t^{n-2}e^t - \int n(n-1)(n-2)t^{n-3}e^t dt \right) \right) \right)$$
Recalling that $\int udv = uv - \int vdu$, we may create a general expression for $n$ terms. Generating the sum, it should be fairly obvious that
$$I = \frac{1}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}t^{n-m}e^t - \int e^t dt \right)$$
Imagine that we have $n!t^0 e^t$ inside of the integral. We may factor out $n!$ and $t^0 = 1$.
We evaluate the integral, and then factor out $e^t$
$$ I = \frac{e^t}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}t^{n-m} - 1 \right) + C$$
And we substitute $t=nx$ back in
$$ I = \frac{e^{nx}}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}(nx)^{n-m} - 1 \right) + C$$
Where did I go wrong? It seems to particularly be the $\frac{1}{n^{n+1}}$ factor that messes up results. I tried manually evaluating for n = 1, 2, 3, 4 and compared to calculators.

 

[jedishrfu]

This may help its a bit more general replacing nx with ax but should still give the same answer:



Be careful, if you make a mistake again the instructor is likely to bench press you. :-)

 

[Svein]

Use partial integration: $\int (x^{n} e^{nx})dx=x^{n}\frac{1}{n} e^{nx}-\frac{1}{n} \int (nx^{n-1}e^{nx})dx$. Sooner or later you will end up in $\int (x^{0} e^{nx})dx$.

Or you can start at $\int (x e^{nx})dx$ and work your way up using induction.

 

[Adesh]

Notice that you got a product of two functions, $x^n$ and $e^nx$. So, from elementary classes we can say that integration by parts will be the best method here.

 

[Mayhem]

Use partial integration: $\int (x^{n} e^{nx})dx=x^{n}\frac{1}{n} e^{nx}-\frac{1}{n} \int (nx^{n-1}e^{nx})dx$. Sooner or later you will end up in $\int (x^{0} e^{nx})dx$.

Or you can start at $\int (x e^{nx})dx$ and work your way up using induction.

So, instead of doing a substitution, I can should do IBP from the get-go?

 

[Mayhem]

So it seems I was ONLY wrong by the -1 term, and after having slept on it, I see my mistake. It would only be true if the bound of the sum was n-1, but that would be redundant as letting the bound equal n gives us

$$ I = \frac{e^{nx}}{n^{n+1}}\left(\sum_{m = 0}^{n} \frac{(-1)^mn!}{(n-m)!}(nx)^{n-m}\right) + C$$

which concludes the evaluation. :)