Integral of $\int z^3 e^z$

Question

I am not sure how I am suppose to do this so I set it up normally.

$\int z^3 e^z$

$u = z^3 dx$ $du = 3z^2$

$dv = e^z$ $v = e^z$

$e^z z^3 - \int e^z 3z^2$

This then leads to a loop of integration by part and I am not sure what to do.

Answer 1

It does not lead to an infinite loop: the exponent on $z$ was reduced by $1$, and if you repeat the process two more times, it will disappear, leaving you with a simple exponential to integrate.

However, what you’ve written is incorrect, starting with the missing $dz$ in the integral: $\int z^3e^z\,\underline{dz}$. For your integration by parts $u=z^3$, not $z^3\,dz$, and $dv=e^zdz$, not $e^z$: differentials can only be equal to other differentials. Similarly, $du=3z^2\underline{dz}$, and $v=e^z$. And there is no $dx$ in the problem anywhere.

Answer 2

$$I_n = \int z^n \exp(z) dz = \int z^n d (\exp(z)) = z^n \exp(z) - n \int z^{n-1} \exp(z) dz = z^n \exp(z) - n I_{n-1}$$ Note that $I_0 = \exp(z) + C$. Hence, $$I_1 = z \exp(z) - \exp(z) +C_1$$ Make use of the same idea to compute $I_2$ and $I_3$.

Move your mouse over the gray area for the answer.

$$I_2 = z^2 \exp(z) - 2 \left(z \exp(z) - \exp(z) \right) + C_2$$ $$I_3 = z^3 \exp(z) - 3 \left(z^2 \exp(z) - 2 \left(z \exp(z) - \exp(z) \right) \right) + C_2$$ and so on$\ldots$

Answer 3

Their is a class of functions, $f(x)=g(x)*h(x)$, where the $g(x)$ has it's $n^{th}$ derivative being zero. One integration by parts will give you

1). $\int h(x)g(x)dx=H(x)g(x)-\int H(x)g'(x)dx$ where $H(x)$ is an antiderivative of $g(x)$. , where $H(x)$ is an antiderivative of $h(x)$.

But then the integral

2) $\int H(x)g'(x)dx=H_2(x)g'(x)-\int H_2(x)g''(x)dx$. where $H_2(x)$ is an antiderivative of $H(x)$.

By plugging (2) into (1), we get

3) $\int h(x)g(x)dx=H(x)g(x)-H_2(x)g'(x)+\int H_2(x)g''(x)dx$.

By repeating this prossess, we get a formula

4) $\int h(x)g(x)dx=H(x)g(x)-H_2g'+H_3g''-H_4g^{(3)}+\ldots$

With all of this, we can organize this prossess into an easily understandable and usable table (this is called the tabular method, it is nothing more than reapeated integration by parts). Let me show you what I mean with the example in your question, $\int z^3e^zdz$. Note that if you continue to differentiate $z^3$, you will get zero. So lets make a table;

$+-+-+-$ (alternaing +/-)

$z^3,3z^2,6z,6,0$ (the derivatives of $z^3$)

$e^z,e^z,e^z,e^z,e^z$ (anitderivatives of $e^z$)

When you put this together, you get,

$\int z^3e^zdz=e^zz^3-3z^2e^z+6ze^z-6e^z+c$

Just to round out the disscussion, lets do another example, $\int \sin(z)z^2dz$. Making the analogous table,

$+-+-+$ (alternaing +/-)

$z^2,2z,2,0$ (the derivatives of $z^2$)

$\sin(z),-\cos(z),-\sin(z),\cos(z)$ (anitderivatives of $\sin(z)$

Putting this together in the same way as before, we get

$\int \sin(z)z^2dz=(z^2)*(-\cos(z))-(2z)*(-\sin(z))+(2)(\cos(z))+c=-z^2\cos(z)+2z\sin(z)+\cos(z)+c$

Also, here is a nice youtube video

http://www.youtube.com/watch?v=zVIE1N156-0

Answer 4

Continue doing integration by parts until the there is no more z.