Tjyylli

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

$$intfrac{dt}{(t-2)(t+3)}$$

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.

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_kprod_{j=1}^n (x-a_j)}{x-a_k}
=sum_{k=1}^n b_kprod_{j=1, jne 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_iprod_{j=1, jne i}^n (a_i-a_j)
$
so that
$b_i
=dfrac1{prod_{j=1, jne 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)}
$.

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 $$A_1 = frac {1}{(1-3)(1+4)}=frac{-1}{10}$$
$$A_2 = frac {1}{(3-1)(3+4)} = frac{1}{14}$$

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

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 $-alpha$:

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

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

Let $n$,$m$ be integers such that $nge 2$ 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}{prodlimits_{j=1}^n (x-b_j)} = sumlimits_{i=1}^n frac{1}{x-b_i} cdot frac{b_i^m}{prodlimits_{j=1,jneq 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 $mge 0$ the following formula holds true:
begin{eqnarray}
frac{x^m}{(x-b_1)^{m_1} (x-b_2)^{m_2}}=sumlimits_{j=0}^m left( sumlimits_{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}}+
sumlimits_{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 $n > 2$.