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Below is a problem from the book "Calculus and Analytic Geometry" by Thomas Finney. I am hoping somebody can check my work. I consider it to be a particular hard problem. Thanks,
Bob

Problem:

We dare you to evaluate this integral.
$$ \int \frac{1}{x(x+1)(x+2)(x+3)(x+4) ... (x+m)} \,\, dx $$

Answer:
To evaluate this integral, we will consider some special cases. For $m = 0$ we have: \begin{align*} \int \frac{1}{x} \,\, dx &= \ln|x| + C \\ \end{align*}


Now for $m = 1$ we have the following integral: $$ \int \frac{1}{x(x+1)} \,\, dx $$ \begin{align*} \frac{1}{x(x+1)} &= \frac{A}{x} + \frac{B}{x+1} \\ 1 &= A(x+1) + B(x) \\ \end{align*} At $x = 0$ we have $1 = A(0+1)$ which yields $A = 1$. \begin{align*} A + B &= 0 \\ 1 + B &= 0 \\ B &= -1 \\ \frac{1}{x(x+1)} &= \frac{1}{x} - \frac{1}{x+1} \\ \int \frac{1}{x(x+1)} \,\, dx &= \ln|x| - \ln|x+1| + C \\ \end{align*}


Now for $m = 2$ we have the following integral: $$ \int \frac{1}{x(x+1)(x+2)} \,\, dx $$ \begin{align*} \frac{1}{x(x+1)(x+2)} &= \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2} \\ 1 &= A(x+1)(x+2) + B(x)(x+2) + C(x)(x+1) \\ \end{align*}

  • At $x = 0$ we have $1 = A(0+1)(0+2)$ which yields $A = \frac{1}{2}$.
  • At $x = -1$ we have $1 = B(-1)(-1+2) = -B$ which yields $B = -1$.
  • At $x = -2$ we have $1 = C(-2)(-2+1) = 2C$ which yields $C = \frac{1}{2}$. \begin{align*} \frac{1}{x(x+1)(x+2)} &= \frac{ \frac{1}{2}}{x} - \frac{1}{x+1} + \frac{ \frac{1}{2}}{x+2} \\ \int \frac{1}{x(x+1)(x+2)} \,\, dx &= \frac{1}{2} \ln{|x|} - \ln{|x+1|} + \frac{1}{2} \ln{|x+2|} + C \\ \end{align*}

Now for $m = 3$ we have the following integral: $$ \int \frac{1}{x(x+1)(x+2)(x+3)} \,\, dx $$ \begin{align*} \frac{1}{x(x+1)(x+2)(x+3)} &= \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2} + \frac{D}{x+3} \\ 1 &= A(x+1)(x+2)(x+3) + B(x)(x+2)(x+3) + \\ & C(x)(x+1)(x+3) + D(x)(x+1)(x+2) \\ \end{align*}

  • At $x = 0$ we have $1 = A(0+1)(0+2)(0+3) = 6A$ which yields $A = 1/6$.
  • At $x = -1$ we have $1 = B(-1)(-1+2)(-1+3) = -2B$ which yields $B = -1/2$.
  • At $x = -2$ we have $1 = C(-2)(-2+1)(-2+3) = 2C$ which yields $C = 1/2$.
  • At $x = -3$ we have $1 = D(-3)(-3+1)(-3+2) = -6D$ which yields $D = -1/6$.

Hence, we have the following solution: $$ \int \frac{1}{x(x+1)(x+2)(x+3)} \,\, dx = \frac{1}{6}\ln{|x|} - \frac{1}{2}\ln{|x+1|} + \frac{1}{2}\ln{|x+2|} - \frac{1}{6}\ln{|x+3|} + C $$


Now for $m = 4$ we have the following integral: $$ \int \frac{1}{x(x+1)(x+2)(x+3)(x+4)} \,\, dx $$ \begin{align*} \frac{1}{x(x+1)(x+2)(x+3)(x+4)} &= \frac{A}{x} + \frac{B}{x+1} + \frac{C}{x+2} + \frac{D}{x+3} + \frac{E}{x+4} \\ 1 &= A(x+1)(x+2)(x+3)(x+4) \\ &+ B(x)(x+2)(x+3)(x+4) \\ &+ C(x)(x+1)(x+3)(x+4) \\ &+ D(x)(x+1)(x+2)(x+4) \\ &+ E(x)(x+1)(x+2)(x+3) \\ \end{align*}

  • At $x = 0$ we have $1 = A(0+1)(0+2)(0+3)(0+4) = 24A$ which yields $A = 1/24$.
  • At $x = -1$ we have $1 = B(-1)(-1+2)(-1+3)(-1+4) = -6B$ which yields $B=-1/6$.
  • At $x = -2$ we have $1 = C(-2)(-2+1)(-2+3)(-2+4) = 4C$ which yields $C = 1/4$.
  • At $x = -3$ we have $1 = D(-3)(-3+1)(-3+2)(-3+4) = -6D$ which yields $D=-1/6$.
  • At $x = -4$ we have $1 = E(-4)(-4+1)(-4+2)(-4+3) = 24E$ which yields $E=1/24$.

Hence we have the solution:
Now for $m = 4$ we have the following integral: $$ \int \frac{1}{x(x+1)(x+2)(x+3)(x+4)} \,\, dx = \frac{\ln{|x|} - 4\ln{|x+1|} + 6\ln{|x+2|} - 4\ln{|x+3|} + \ln{|x+4|}}{24} + C $$


Now let's consider the general case. \begin{align*} \frac{1}{x(x+1) \cdots (x+m)} \,\, &= \frac{C_0}{x} + \frac{C_1}{x+1} \cdots + \frac{C_m}{x+m} \\ \end{align*} \begin{align*} 1 &= {C_0}(x+1)(x+2) \cdots (x+m) \\ &+ {C_1}(x)(x+2) \cdots (x+m) \\ &+ {C_2}(x)(x+1)(x+3)(x+4) \cdots (x+m) \\ &+ \cdots \\ &+ C_m(x)(x+1)(x+2) \cdots (x+m-1) \\ \end{align*} Now let's consider the first term. We set $x = 0$ and we get: \begin{align*} 1 &= {C_0}(0+1)(0+2) \cdots (x+m) = m! C_0 \\ C_0 &= \frac{1}{m!} \end{align*} Now let's consider the $C_2$ term. We set $x = 2$ and $m > 4$. We get: \begin{align*} 1 &= C_2(-2)(-2+1)(-2+3)(-2+4)(-2 + 5) \cdots (-2 + m) \\ 1 &= C_2 (2)(1)(2)(3)(4) \cdots (m-2) \\ 1 &= 2(m-2)! C_2 \\ C_2 &= \frac{1}{2(m-2)!} = \frac{m(m-1)}{2(m!)} \\ C_2 &= \frac{ \binom {m}{2} } {m!} \\ \end{align*} Now let's consider the last term. We set $x = -m$ and we get:
\begin{align*} 1 &= C_m(-m)(-m+1)(-m+2) \cdots (x+m -1) \\ C_m &= \frac{(-1)^{m}}{m!} \\ \end{align*} Now let's consider one of the middle terms. We set $x = -k$ where $0 <= k <= m$ and we get: \begin{align*} 1 &= C_k(-k)(-k+1)(-k+2) \cdots (-1) (1)(2) \cdots (-k + m - 1) \\ 1 &= {-1}^k C_k(k-1)(k-2) \cdots (1)(2) \cdots (-k + m - 1) \\ C_k &= \frac{ 1 }{ {(-1)}^k C_k(k-1)(k-2) \cdots (1)(2) \cdots (-k + m - 1) } \\ C_k &= \frac{ k! }{ {(-1)}^k C_k(k-1)(k-2) \cdots (1)m! } \\ C_k &= \frac{ \binom {m}{k} }{ {(-1)}^k m! } \\ \end{align*} Hence the answer is: $$ \sum_{k=0}^{k=m} \left( \frac{ \binom {m}{k} }{ {(-1)}^k m! }\right) \ln{|x+k|} + C $$

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    $\begingroup$ What's your question? $\endgroup$
    – user376343
    May 19, 2019 at 19:05
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    $\begingroup$ @user376343 - From the opening of the post, "I am hoping somebody can check my work." $\endgroup$ May 19, 2019 at 19:07
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    $\begingroup$ It seems it is right, but the powers of $-1$ must be in parentheses. ... and it is usual to put them in numerators rather than in denominators. $\endgroup$
    – user376343
    May 19, 2019 at 19:12
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    $\begingroup$ There are no mistakes in your work per se. However, you should try to make solutions to problems as concise as possible without sacrificing clarity. If I were teaching a class how to do this problem, I would probably show all the work you showed in your answer. If I were doing this problem for homework, I would set up a standard induction argument. Base cases ($m = 0$, $m = 1$): $\ldots$. Inductive step: $\ldots$. Your prof probably doesn't want to see your thought process; he just wants to see a clear and complete solution. $\endgroup$ May 19, 2019 at 19:57
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    $\begingroup$ @Bob the powers of $−1$ must be in parentheses, otherwise it is always $-1$. $\endgroup$
    – user376343
    May 19, 2019 at 20:20

6 Answers 6

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A simpler approach is possible without all the other preliminary work. Let $$q_m(x) = \prod_{k=0}^m (x+k), \quad f_m(x) = \frac{1}{q_m(x)}.$$ Then $f$ admits a partial fraction decomposition of the form $$f_m(x) = \sum_{n=0}^m \frac{A_n}{x+n} \tag{1}$$ for suitable constants $A_0, \ldots, A_m$ which we wish to find, hence an antiderivative of $f$ is $$\int f_m(x) \, dx = \sum_{n=0}^m A_n \log |x+n|. \tag{2}$$ (I have omitted the constant of integration for convenience.) So all that remains is to determine the form of $A_n$. To do this, we observe that $$1 = q_m(x) \sum_{n=0}^m \frac{A_n}{x+n} = \sum_{n=0}^m p_n(x) A_n,$$ where $$p_n(x) = \prod_{k \ne n} (x+k) = (-1)^n \prod_{k=0}^{n-1} (-x-k) \prod_{k=n+1}^m (k + x).$$ Then in particular $$p_n(-n) = (-1)^n \prod_{k=0}^{n-1} (n-k) \prod_{k=n+1}^m (k-n) = (-1)^n n!(m-n)! = \frac{(-1)^n m!}{\binom{m}{n}},$$ and $p_n(-k) = 0$ for all other nonnegative integers $k \le m$ not equal to $n$. Therefore, $$A_n = \frac{1}{p_n(-n)} = \frac{(-1)^n}{m!} \binom{m}{n}$$ and $$\int f_m(x) \, dx = \sum_{n=0}^m \frac{(-1)^n}{m!} \binom{m}{n} \log |x+n| + C$$ as claimed.

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At this line: "Now let's consider the first term. We set $x=0$ and we get:" $$1 = {C_0}(0+1)(0+2) \cdots (x+m) = m!C_0$$ When you set $x=0$ there won't be any $x$ left.

For $C_2$ there is no mistake but then for the general term you did the same typo.

You can't have after you set $x=-m$: $$1 = C_m(-m)(-m+1)(-m+2) \cdots (\color{red}x+m -1)$$ And for the last part for some reason, you have $C_k$ in the denominator. $$\begin{align*} 1 &= C_k(-k)(-k+1)(-k+2) \cdots (-1) (1)(2) \cdots (-k + m - 1) \\ 1 &= (-1)^k C_k(k-1)(k-2) \cdots (1)(2) \cdots (-k + m - 1) \\ C_k &= \frac{ 1 }{ (-1)^k \color{red}{C_k}(k-1)(k-2) \cdots (1)(2) \cdots (-k + m - 1) } \\ C_k &= \frac{ k! }{ (-1)^k \color{red}{C_k}(k-1)(k-2) \cdots (1)m! } \\ C_k &= \frac{ \binom {m}{k} }{ (-1)^k m! } \\ \end{align*}$$ Other than this everything it's correct.

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    $\begingroup$ Minor nitpick. It's more appropriate to have $(-1)^k$ instead of $-1^k$. It makes it clearer that it's $-1$ being raised to the power. Otherwise per the order of operations, for example, $$-2^6 = -(2^6) = -64 \ne (-2)^6 = 64$$ $\endgroup$ May 20, 2019 at 2:07
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    $\begingroup$ @EeveeTrainer Actually it was a LaTeX typo: {-1}^k instead of (-1)^k. $\endgroup$ May 20, 2019 at 7:13
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We could set up a difference equation. Partial fractions indicates that $$f_m(x)=\frac1{\prod_{k=0}^m(x+k)}=\sum_{k=0}^m\frac{A_k^{(m)}}{x+m}$$ And it is a simple calculation to show that $$f_{m-1}(x)-f_{m-1}(x+1)=mf_m(x)$$ So, comparing coefficients of $\frac1{x+k}$ we have $A_0^{(m-1)}=mA_0^{(m)}$, $A_{m-1}^{(m-1)}=-mA_m^{(m)}$, and $A_k^{(m-1)}-A_{k-1}^{(m-1)}=mA_k^{(m)}$ for $1\le k\le m-1$. If we let $A_k^{(m)}=\frac{(-1)^k}{m!}B_k^{(m)}$ then our difference equations read $B_0^{(m-1)}=B_0^{(m)}=\cdots=B_0^{(0)}=A_0^{(0)}=1$, $B_{m-1}^{(m-1)}=B_m^{(m)}=\cdots=B_0^{(0)}=1$ and $B_k^{(m-1)}+B_{k-1}^{(m-1)}=B_k^{(m)}$ for $1\le k\le m-1$. We recognize these as the difference equations for Pascal's triangle, so $B_k^{(m)}={m\choose k}$ so it follows that $A_k^{(m)}=\frac{(-1)^k}{m!}{m\choose k}$ and $$\int f_m(x)dx=\sum_{k=0}^m\frac{(-1)^k}{m!}{m\choose k}\ln|x+k|+C$$

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Another way:

We have:

$$ \frac{1}{x (x+1)...(x+m)} = \sum_{j=0}^{j=m} \frac{a_j}{x+j}$$

By the Heaviside coverup method,

$$ a_j = \lim_{x \to (-j)} \frac{x+j}{x (x+1)...(x+m)}$$

Compute a few values of $a_j$:

$$ a_1 = \frac{1}{(-1) (2) (3)...(m)}= \frac{(-1)^1}{(m-1)!}$$

$$a_2 = \frac{1}{ (-2) (-1) (1) (2)... (m-2)!} =\frac{(-1)^2}{2!(m-2)!}$$

$$a_3 = \frac{}{(-3)(-2)(-1)(1)...(m-3)} = \frac{(-1)^3}{3! (m-3)!}$$

By induction,

$$ a_j = \frac{(-1)^j}{j! (m-j)!} = \frac{(-1)^j}{m!} \binom{m}{j}$$

Now integrate both sides in the first expression,

$$ \int \frac{1}{x (x+1)...(x+m)} dx = \sum_{j=0}^{j=m} \int \frac{a_j}{x+j} dx $$

Or,

$$ \int \frac{1}{x (x+1)...(x+m)} dx = \sum_{j=0}^{j=m} a_j \ln( |x+j|) +C= \sum_{j=0}^{j=m} \frac{(-1)^j}{m!} \binom{m}{j} \ln( |x+j|) +C $$

Ultimately,

$$\int \frac{1}{x (x+1)...(x+m)} dx = \frac{1}{m!} \sum_{j=0}^{j=m}(-1)^j \binom{m}{j} \ln( |x+j|) +C $$

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Let us use the identity that $$f(x)=\frac{1}{(x+1)(x+2)(x+3)...(x+n)}=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k+1} \frac{k{n \choose k}}{x+k}.$$ Let $$g(x)=\frac{1}{x(x+1)(x+2)(x+3)...(x+n)}=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k+1} \frac{k{n \choose k}}{x(x+k)}.$$ $$\implies g(x)=\frac{1}{n!}\sum_{k=1}^{n} (-1)^{k+1} {n \choose k} \left(\frac{1}{x}-\frac{1}{x+k}\right)$$ $$\implies g(x)=\frac{1}{n! x}-\frac{1}{n!}\sum_{k=1}^{n}(-1)^{k+1} \frac{{n \choose k}}{x+k}.$$ Finally, $$\int g(x) dx=\frac{1}{n!}\left(\ln x+\sum_{k=1}^{n} (-1)^{k} {n \choose k} \ln(x+k)\right)$$

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It looks right to me [That is, I did the calculation independently and got the same answer]. For brevity/clarity purposes you really only need to include the “general case”.

It would be more conventional to write $$ C_k = \frac{(-1)^k}{k!(m-k)!} $$

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