What is the indefinite integral of 1/(x(x+1)(x+2)...(x+m))dx?

[member 508213]

Homework Statement

Indefinite integral of 1/(x(x+1)(x+2)...(x+m))dx

Homework Equations

None

The Attempt at a Solution

The problem seems to be a conceptual one and naturally in a format like that I think partial fractions expansion... However that does not seem to work here.. Thanks for any help I am lost on where to start

 

[vela]

Why doesn't partial fractions work?

 

[member 508213]

Why doesn't partial fractions work?

I thought since there are infinite terms in the denominator it would never happen?

 

[vela]

The way you wrote it, it appears to be a finite number of factors in the denominator. Is it supposed to be infinite?

 

[member 508213]

The way you wrote it, it appears to be a finite number of factors in the denominator. Is it supposed to be infinite?

I may not have full understanding of the question because when I saw the "..." Followed by the x + arbitrary m, i assumed that meant infinite or atleast undefined. Do you see it differently?

 

[member 508213]

The question is written in the textbook the way I wrote it in the question just to clarify, with the "..."

 

[vela]

That would mean there are m+1 factors: x, x+1, ..., x+(m-1), and x+m. The result will depend on m, obviously. It's like when one writes $n! = 1\times2\times3\times\cdots\times n$. Just because a specific value for $n$ isn't given, you don't assume there's an infinite number of factors.

Please note that I'm not saying partial fractions is the best way to do this problem, but without knowing what you've been studying, I think it's a reasonable approach.

 

[member 508213]

That would mean there are m+1 factors: x, x+1, ..., x+(m-1), and x+m. The result will depend on m, obviously. It's like when one writes $n! = 1\times2\times3\times\cdots\times n$. Just because a specific value for $n$ isn't given, you don't assume there's an infinite number of factors.

Please note that I'm not saying partial fractions is the best way to do this problem, but without knowing what you've been studying, I think it's a reasonable approach.

I am in calc bc this problem is supposed to be a difficult college level problem... I understand what you mean that it doesn't mean infinity but it is undefined so i can't see how i could apply partial fractions to this??

 

[member 508213]

And if partial fractions won't work then what can I do??

 

[vela]

I wouldn't say $m$ is undefined. You can assume it's a positive integer. It might help to try work out the partial fraction expansion for a specific value of $m$, like m=5 for example, and see if a pattern emerges that you can generalize for arbitrary $m$. Finally, I suggest you use the Heaviside cover-up method to solve for the various constants.

EDIT: Here's the expansion for $m=5$ from Mathematica. It does appear there's some sort of pattern. Can you figure out what it is?
$$\frac{1}{120 x}-\frac{1}{24 (x+1)}+\frac{1}{12 (x+2)}-\frac{1}{12 (x+3)}+\frac{1}{24 (x+4)}-\frac{1}{120 (x+5)}$$

 

[Ray Vickson]

And if partial fractions won't work then what can I do??

For finite m, partial fractions will work, but they won't be particularly pretty. Assume
f(x) \equiv \frac{1}{x(x+1)(x+2) \cdots (x+m)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \frac{A_2}{x+2} + \cdots + \frac{A_m}{x+m}
Then
x f(x) = \frac{1}{(x+1) \cdots (x+m)} = A_0 + \sum_{i=1}^m \frac{A_i x}{x+i}
so setting $x = 0$ gives $1/m! = A_0$. Similarly,
(x+1) f(x) = \frac{1}{x(x+2) \cdots (x+m)} = A_1 + \sum_{i \neq 1}\frac{ A_i (x+1)}{x+i}
so setting $x = -1$ gives
A_1 = \frac{1}{(-1)(-1+2) \cdots (-1+m)} = -\frac{1}{(m-1)!}
Continue like that to get all the $A_i$.

 

[member 508213]

For finite m, partial fractions will work, but they won't be particularly pretty. Assume
f(x) \equiv \frac{1}{x(x+1)(x+2) \cdots (x+m)} = \frac{A_0}{x} + \frac{A_1}{x+1} + \frac{A_2}{x+2} + \cdots + \frac{A_m}{x+m}
Then
x f(x) = \frac{1}{(x+1) \cdots (x+m)} = A_0 + \sum_{i=1}^m \frac{A_i x}{x+i}
so setting $x = 0$ gives $1/m! = A_0$. Similarly,
(x+1) f(x) = \frac{1}{x(x+2) \cdots (x+m)} = A_1 + \sum_{i \neq 1}\frac{ A_i (x+1)}{x+i}
so setting $x = -1$ gives
A_1 = \frac{1}{(-1)(-1+2) \cdots (-1+m)} = -\frac{1}{(m-1)!}
Continue like that to get all the $A_i$.

I feel like this explanation is good however I have not had much experience with summations so it is difficult for me to fully understand your explanation. I am beginning to see the pattern but I am not sure how I would express this so I would be able to take the integral. I am trying very hard to understand what you are saying but I just haven't had much experience with problems like this so I am having trouble putting it all together.

 

[member 508213]

more specifically I am confused about the equation you set up and the process you are using to find the A values

 

[member 508213]

Additionally, since I cannot tell if m is an odd or even integer how would I tell which of the pieces of the partial fraction will be negative?

 

[member 508213]

So far I believe my partial fractions will look something like this:

1/((m!)(x))- 1/(((m-1)!))(x+1)) +1/(((m-2)!)(x+2)) +...+1/(((m-2)!)(x+(m-2))) - 1/(((m-1)!)(x+(m-1))) + 1/((m!)(x+m))

I am not sure if this is getting anywhere?

Sorry for it looking messy I am not sure how to write fractions on this website yet if anyone could tell me?

 

[vela]

Did you try working it out for m=5?

 

[member 508213]

I guess it would not work then because for the 1/x+2 term it should be 12 but with my model i got 6 but i really don't know what to do from here

 

[Ray Vickson]

I guess it would not work then because for the 1/x+2 term it should be 12 but with my model i got 6 but i really don't know what to do from here

You need to work it out systematically, the way I illustrated for the first two terms 1/x and 1/(x+1). Also, trying it out systematically from start to finish for m = 4 and m = 5 will go a long way towards helping you understand the method and result.

 

[Ray Vickson]

I wouldn't say $m$ is undefined. You can assume it's a positive integer. It might help to try work out the partial fraction expansion for a specific value of $m$, like m=5 for example, and see if a pattern emerges that you can generalize for arbitrary $m$. Finally, I suggest you use the Heaviside cover-up method to solve for the various constants.

EDIT: Here's the expansion for $m=5$ from Mathematica. It does appear there's some sort of pattern. Can you figure out what it is?
$$\frac{1}{120 x}-\frac{1}{24 (x+1)}+\frac{1}{12 (x+2)}-\frac{1}{12 (x+3)}+\frac{1}{24 (x+4)}-\frac{1}{120 (x+5)}$$

It is even more revealing to re-write this after dividing and multiplying by 5! = 120:
f(x) = \frac{1}{5!} \left( \frac{1}{x} - \frac{5}{x+1} + \frac{10}{x+2} - \frac{10}{x+3} + \frac{5}{x+4} - \frac{1}{x+5} \right)
The successive coefficients inside ( ) ought to be instantly recognizable.