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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.

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4  
why does it lead to a loop? You reduced $z^3$ to $z^2$ in the integral after all. What happens if you keep doing that? –  Robert Mastragostino May 29 '12 at 23:58
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You should digest the answers to your old questions before you post new ones. I showed you how to set up the integration by parts, and you're still making the $u=z^3\,dx$, $dv=e^z$ mistake, only worse, since there isn't any $x$ in the problem. –  Gerry Myerson May 30 '12 at 0:01
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No loop: $3$, $2$, $1$, $0$, bingo! –  André Nicolas May 30 '12 at 0:13
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It's interesting that the upvotes from the first comment down to Andre's follows a order of $3, 2, 1$ as well! –  Joe May 30 '12 at 3:11

4 Answers 4

up vote 1 down vote accepted

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.

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I do not understand how the exponent is reduced, mine increases. –  user138246 May 30 '12 at 10:08
    
@Jordan: Yours DEcreased: your original integral had an integrand of $z^3e^z$, and your new one has an integrand of $z^2e^z$: the exponent on $z$ is now only $2$, where it was originally $3$. (The constant multiplier of $3$ is not a problem.) Do it again, and you’ll have an integral whose integrand is $ze^z$; do it a third time, and the integrand of the remaining integral will be $e^z$. The exponent on $z$ decreases by $1$ each time. –  Brian M. Scott May 30 '12 at 10:11

$$I_n = \int z^n \exp(z) dz = \int z^n d (\exp(z)) = z^n \exp(z) - n \int z^{n-1} \exp(z) dz = z^n \exp(z) - n I_{n-1}$$ 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$\ldots$

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Fight fat with a reduction formula! It's gonna work. –  ncmathsadist May 30 '12 at 1:02
    
I am just confused by all this notation, exp(z) I $z^n$ and all of that. –  user138246 May 30 '12 at 10:18
    
I have no idea what is happening in this post, I don't think I am this far in calculus yet. –  user138246 Jun 1 '12 at 12:37
    
@Jordan $\exp z = e^z$. Marvis is using the integration by parts formula, with $u=z^n$ and $dv=e^z dz$. Try to do it that way. –  Pedro Tamaroff Jun 6 '12 at 22:17

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

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Honestly this just feels like another trick that I would need to memorize when I feel it would be more beneficial to try and learn just how to do this in a simple way that does not require additional tricks to be memorized. Especially since this seems to be a rare case that hardly makes the problem easier at all. –  user138246 Jun 1 '12 at 12:40
    
I sympathize with your feeling in a way, that this subject seems to be a collection of tricks, and that you have to be a master alchemist. That said, this particular trick is not so much a trick but a way of nicely organize the situation where you have to do repeated integration by parts. –  Baby Dragon Jun 1 '12 at 19:09

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

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1  
With comments suggesting this already and the lack of content this provides, why make this into an answer? –  Joe May 30 '12 at 3:12

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