Solve Integral: $\int \frac{1}{x(x+1)(x+2)(x+3)...(x+m)}dx$

[coki2000]

Hello PF members,
Can you help me to solve this integral please?

\int \frac{1}{x(x+1)(x+2)(x+3)...(x+m)}dx

How can i solve this? Thanks for your helps.

 

[phyzguy]

Use the method of partial fractions.

 

[HallsofIvy]

In other words, find A₀, A₁, A₂, ..., A_m
such that
\frac{1}{x(x-1)(x-2)\cdot\cdot\cdot(x-m)}= \frac{A_0}{x}+ \frac{A_1}{x- 1}+ \frac{A_2}{x- 2}+ \cdot\cdot\cdot+ \frac{A_m}{x- m}

Integrating that will give you a sum of logarithms:
A_0 ln(x)+ A_1 ln(x- 1)+ A_2 ln(x- 2)+ \cdot\cdot\cdot+ A_m ln(x- m)
= ln(x^{A_0}(x- 1)^{A_1}(x- 2)^{A_2}\cdot\cdot\cdot(x- m)^{A_m})

 

[coki2000]

In other words, find A₀, A₁, A₂, ..., A_m
such that
\frac{1}{x(x-1)(x-2)\cdot\cdot\cdot(x-m)}= \frac{A_0}{x}+ \frac{A_1}{x- 1}+ \frac{A_2}{x- 2}+ \cdot\cdot\cdot+ \frac{A_m}{x- m}

Integrating that will give you a sum of logarithms:
A_0 ln(x)+ A_1 ln(x- 1)+ A_2 ln(x- 2)+ \cdot\cdot\cdot+ A_m ln(x- m)
= ln(x^{A_0}(x- 1)^{A_1}(x- 2)^{A_2}\cdot\cdot\cdot(x- m)^{A_m})

Thank you but how can i find the coefficients a1 a2 ...?

 

[HallsofIvy]

?? That is a standard method for integrating rational functions. I find it hard to believe that you are attempting an integral like this without having been introduced to "partial fractions".

There are many different ways to find the coefficients but for this problem the simplest is: from
\frac{1}{x(x-1)(x-2)\cdot\cdot\cdot(x-m)}= \frac{A_0}{x}+ \frac{A_1}{x- 1}+ \frac{A_2}{x- 2}+ \cdot\cdot\cdot+ \frac{A_m}{x- m}
multiply both sides by the denominator on the left. That will cancel all of the denominators to give
1= A_0(x-1)(x-2)\cdot\cdot\cdot(x- m)+ A_1x(x-2)(x-3)\cdot\cdot\cdot(x-m)+ A_2x(x-1)(x-3)\cdot\cdot\cdot(x-m)+ \cdot\cdot\cdot+ A_mx(x- 1)(x- 2)\cdot\cdot\cdot(x- m+1)

Now, take x= 0, 1, 2, ..., m-1, m and all except one of the terms is 0. For example if x= 0 then
1= A_0(-1)(-2)\cdot\cdot\cdot(-m)= (-1)^m m!A_0
and so
A_0= \frac{(-1)^m}{m!}

 

[dimitri151]

Looks like a pretty clear rule for m odd.

Isn't there a way to use the gamma function?