9
$\begingroup$

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 evalaute 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 $A = \frac{1}{2}$. \newline At $x = -1$ we have $1 = B(-1)(-1+2) = -B$ which yields $B = -1$. \newline 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*} \newline 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 $A = \frac{1}{6}$.

At $x = -1$ we have $1 = B(-1)(-1+2)(-1+3) = -2B$ which yields $B = -\frac{1}{2}$.

At $x = -2$ we have $1 = C(-2)(-2+1)(-2+3) = 2C$ which yields $C = \frac{1}{2}$.

At $x = -3$ we have $1 = D(-3)(-3+1)(-3+2) = -6D$ which yields $D = -\frac{1}{6}$.

Hence, we have the following solution: $$ \int \frac{1}{x(x+1)(x+2)(x+3)} \,\, dx = \frac{1}{6}\ln{|x|} - \frac{1}{2}\ln{|x+1|} + \frac{1}{2}\ln{|x+2|} - \frac{1}{6}\ln{|x+3|} + C $$ \newline 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 $A = \frac{1}{24}$.

At $x = -1$ we have $1 = B(-1)(-1+2)(-1+3)(-1+4) = -6B$ which yields $B = -\frac{1}{6}$.

At $x = -2$ we have $1 = C(-2)(-2+1)(-2+3)(-2+4) = 4C$ which yields $C = \frac{1}{4}$.

At $x = -3$ we have $1 = D(-3)(-3+1)(-3+2)(-3+4) = -6D$ which yields $D = -\frac{1}{6}$.

At $x = -4$ we have $1 = E(-4)(-4+1)(-4+2)(-4+3) = 24E$ which yields $E = \frac{1}{24}$.

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)(x+2)(x+3)(x+4) \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 lets 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 lets 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 lets 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 lets 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: $$ \sum_{k=0}^{k=m} \left( \frac{ \binom {m}{k} }{ {(-1)}^k m! }\right) \ln{|x+k|} + C $$

$\endgroup$
  • $\begingroup$ What's your question? $\endgroup$ – user376343 May 19 at 19:05
  • 2
    $\begingroup$ @user376343 - From the opening of the post, "I am hoping somebody can check my work." $\endgroup$ – Eevee Trainer May 19 at 19:07
  • 3
    $\begingroup$ 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 at 19:12
  • 3
    $\begingroup$ 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. $\endgroup$ – Charles Hudgins May 19 at 19:57
  • 2
    $\begingroup$ @Bob the powers of $−1$ must be in parentheses, otherwise it is always $-1$. $\endgroup$ – user376343 May 19 at 20:20
5
$\begingroup$

At this line: "Now lets 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.

$\endgroup$
  • 1
    $\begingroup$ 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, $$-2^6 = -(2^6) = -64 \ne (-2)^6 = 64$$ $\endgroup$ – Eevee Trainer May 20 at 2:07
  • 1
    $\begingroup$ @EeveeTrainer Actually it was a LaTeX typo: {-1}^k instead of (-1)^k. $\endgroup$ – Jean-Claude Arbaut May 20 at 7:13
11
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $$\int f_m(x)dx=\sum_{k=0}^m\frac{(-1)^k}{m!}{m\choose k}\ln|x+k|+C$$

$\endgroup$
0
$\begingroup$

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 $$ C_k = \frac{(-1)^k}{k!(m-k)!} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.