7

Often I am dealing with an integral of let's say:

$$\int\frac{dt}{(t-2)(t+3)}$$

or

$$\int \frac{dt}{t(t-4)}$$

or to make this a more general case in which I am interested the most:

$$\int \frac{dt}{(t+\alpha)(t+\beta)} \quad \quad \alpha, \beta \in \mathbb{R}$$

Basically any integral with decomposed polynomial of second degree in the denominator. Obviously this leads to sum of two natural logarithms.

What I do is write down partial fractions equation and then solve system of linear equations (2 variables here):

$$\frac{1}{(t+\alpha)(t+\beta)} = \frac{A}{t+\alpha} + \frac{B}{t+\beta}$$

After solving this I end up with some $A, B$ coefficients and I can solve the integral.

Is there faster way to find $A, B$ ? Some algorithm or anything that I can follow and would always work for such case? Surely, solving that does not take much time but I just wonder if it could be done even faster.

(bonus question: what if there were more variables, like 3 variables?)

I would greatly appreciate all feedback as it could help me save countless number of minutes in the future.

weno
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  • Also when I think about it, I wonder if there's even a general method (beyond the one you referred to). I can see very easily a general solution for when (i) the numerator is $1$ (ii) $\alpha \ne \beta$ (iii) we only have a product of two factors in the denominator, but that's such a narrow set of circumstances that I wonder if it's worth it. Maybe someone has a more general form though. – PrincessEev Apr 12 '19 at 01:25

4 Answers4

9

If your fraction is in form of $$ \frac {1}{(t-\alpha_1)(t-\alpha_2)(t-\alpha_3)...(t-\alpha_k)}$$ where the $\alpha s$ are different, there is an easy way to find the partial fraction decomposition.

Let $t=\alpha_i$ and cover $(t-\alpha_i)$ while evaluating the above fraction to get your $A_i$

For example $$ \frac{1}{(t-1)(t-3)(t+4)} = \frac {A_1}{(t-1)} + \frac{ A_2}{(t-3)} +\frac { A_3}{(t+4)}$$ Where $$A_1 = \frac {1}{(1-3)(1+4)}=\frac{-1}{10}$$ $$A_2 = \frac {1}{(3-1)(3+4)} = \frac{1}{14}$$

$$A_3=\frac {1}{(-4-1)(-4-3)}=\frac{1}{35}$$

Mohammad Riazi-Kermani
  • 70,421
7

Unless I am misunderstanding the question, by "root rotation" (when $\beta \neq \alpha$):

$$\frac{1}{(t + \alpha)(t + \beta)} = \frac{A}{t + \alpha} + \frac{B}{t + \beta}$$

$$1 = A(t + \beta) + B(t + \alpha)$$

Evaluating $-\beta$ for $t$:

$$1 = B(\alpha - \beta)$$

$$B = \frac{1}{\alpha - \beta}$$

Similarly, for $A$, sub in $-\alpha$:

$$1 = A(\beta - \alpha)$$

$$A = \frac{1}{\beta - \alpha}$$

MichaelChirico
  • 4,271
Dair
  • 3,654
  • I'll be coming back to this post. This is what I was looking for. – weno Apr 12 '19 at 01:45
  • Unless that takes you too much time, would you be able to derive formulas for $A, B, C$, please? Function of form $\frac{1}{(t+\alpha)(t+\beta)(t+\gamma)}$ and $\alpha \neq \beta \neq \gamma$ – weno Apr 12 '19 at 01:47
6

Here's your answer for general $n$.

$\dfrac1{\prod_{k=1}^n (x-a_k)} =\sum_{k=1}^n \dfrac{b_k}{x-a_k} $.

Therefore $1 =\sum_{k=1}^n \dfrac{b_k\prod_{j=1}^n (x-a_j)}{x-a_k} =\sum_{k=1}^n b_k\prod_{j=1, j\ne k}^n (x-a_j) $.

Setting $x = a_i$ for each $i$, all the terms except the one with $k=i$ have the factor $a_i-a_i$, so $1 = b_i\prod_{j=1, j\ne i}^n (a_i-a_j) $ so that $b_i =\dfrac1{\prod_{j=1, j\ne i}^n (a_i-a_j)} $.

For $n=2$, $b_1 =\dfrac1{a_1-a_2} $, $b_2 =\dfrac1{a_2-a_1} $.

For $n=3$, $b_1 =\dfrac1{(a_1-a_2)(a_1-a_3)} $, $b_2 =\dfrac1{(a_2-a_1)(a_2-a_3)} $, $b_3 =\dfrac1{(a_3-a_1)(a_3-a_2)} $.

marty cohen
  • 110,916
1

Let $n$,$m$ be integers such that $n\ge 2$ and $n-1 \ge m \ge 0$ .Then if the roots of the polynomial in question are all distinct the following formula holds: \begin{equation} \frac{x^m}{\prod\limits_{j=1}^n (x-b_j)} = \sum\limits_{i=1}^n \frac{1}{x-b_i} \cdot \frac{b_i^m}{\prod\limits_{j=1,j\neq i}^n (b_i-b_j)} \end{equation} Now, if the roots are not distinct then it gets more complicated but even at the outset it is clear that we can generate a similar formula by differentiating both sides with respect to the particular root a certain number of times. For example if $n=2$ and $m_1 \ge 1$, $m_2 \ge 1$ and $m\ge 0$ the following formula holds true: \begin{eqnarray} \frac{x^m}{(x-b_1)^{m_1} (x-b_2)^{m_2}}=\sum\limits_{j=0}^m \left( \sum\limits_{l_1=1}^{m_1} \binom{m_1+m_2-1-l_1}{m_2-1}(-1)^{m_2} b_1^j \frac{1_{l_1=1} \binom{m-1}{j-1} + 1_{l_1>1} \binom{m}{j}}{(x-b_1)^{l_1-m+j} (-b_1+b_2)^{m_1+m_2-l_1}}+ \sum\limits_{l_1=1}^{m_2} \binom{m_1+m_2-1-l_1}{m_1-1}(-1)^{m_1} b_2^j \frac{1_{l_1=1} \binom{m-1}{j-1} + 1_{l_1>1} \binom{m}{j}}{(x-b_2)^{l_1-m+j} (-b_2+b_1)^{m_1+m_2-l_1}} \right) \end{eqnarray}

In[5732]:= ll = {}; x =.; b1 =.; b2 =.; m = RandomInteger[{0, 10}];
For[count = 1, count <= 100, count++,
  {m1, m2} = RandomInteger[{1, 5}, 2]; x =.; b1 =.; b2 =.;
  xx1 =
      Sum[ 
     Binomial[m1 + m2 - 1 - l1, m2 - 1] ((-1)^
       m2 b1^j If[l1 == 1, Binomial[m - 1, j - 1], 
        Binomial[m, j]])/((x - b1)^(l1 - m + j) (-b1 + b2)^(
       m1 + m2 - l1)), {l1, 1, m1}, {j, 0, m}] + 
    Sum[ Binomial[m1 + m2 - 1 - l1, m1 - 1] ((-1)^
       m1 b2^j If[l1 == 1, Binomial[m - 1, j - 1], 
        Binomial[m, j]])/((x - b2)^(l1 - m + j) (-b2 + b1)^(
       m1 + m2 - l1)), {l1, 1, m2}, {j, 0, m}];
  xx2 = x^m/((x - b1)^m1 (x - b2)^m2);
  ll = Join[ll, {Simplify[xx1 - xx2]}];
  ];
ll

Out[5734]= {0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, \
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0}

It would be interesting to derive similar formulae in the case $n > 2$.

Przemo
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