By trying to find a marginal distribution I came accross integration of the product series. For the sake of generality, lets assume the integral is of following form: $$\int \prod_{k=1}^{n}\left ( x+a_{k} \right )^{b_{k}}dx.$$ $a_{k}$ is a real coefficient and $b_{k}$ is positive integers. Is there any method, that could be used to integrate this analyticaly? It is of no problem to calculate it by some numerical method. But the first thing I would like to try is to find its analytical expression.
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2$\begingroup$ And $b_k$ are what? Positive integers? If so, answer is a polynomial. Integers not necessarily positive? Use partial fractions. In most other cases, closed-form expressions are not likely. $\endgroup$– Robert IsraelCommented Sep 21, 2012 at 19:35
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1$\begingroup$ What do you mean in the question by 'a series of products'? You may already know that each term $(x+a_k)^{b_k}$ may be expanded as a binomial series which has a positive but finite radius of convergence if $b_k$ is not a positive integer. These series may be multiplied, and then you can integrate term by term to get a series expansion of your antiderivative. $\endgroup$– StoppleCommented Sep 21, 2012 at 19:54
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$\begingroup$ What are the $b_k$ and what are the limits of integration? $\endgroup$– Alexandre EremenkoCommented Sep 22, 2012 at 1:06
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2 Answers
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Since the derivative of such a product has a similar form $$\left(\sum_{k=1}^n\frac{b_k}{x+a_k} \right)\prod_{k=1}^{n}\left ( x+a_{k} \right )^{b_{k}},$$ One might hope that something similar is true of some antiderivative. (I am implicitly assuming that the $b_k$ are positive integers, although the formula is valid more generally. However there are cases that will involve some $\ln |x+a_k|$ . I will continue to neglect these.)
If we imagine that we have some one antiderivative $$F(x)=\int \prod_{k=1}^{n}\left ( x+a_{k} \right )^{b_{k}}dx$$ and let $F_k=F(x)-F(-a_k),$ then $F_k(x)$ is also an antiderivative and is divisible by $(x+a_k)^{b_k+1}$
That is not particularly deep.
We can use integration by parts to obtain an answer without any explicit multiplying although marching through to the end will not be very pretty. It is more satisfying to know that we can do it than to actually do so.
- As we know, $\int(x+a_1)^{b_1}dx=\frac{1}{b_{1}+1}(x+a_1)^{b_1+1}$
- We can get to the above form by $b_2$ iterations of $\int(x+a_1)^{b_1}(x+a_2)^{b_2}dx=\frac{1}{b_{1}+1}\left((x+a_1)^{b_1+1}(x+a_2)^{b_2}-b_2\int(x+a_1)^{b_1+1}(x+a_2)^{b_2-1}dx\right)$
- Similarly, we can eventually get from three factors to two etc.
It would be only slightly more complicated (but also only slightly more satisfying) to have products of factors $(m_kx+a_k)^{b_k}$.
A large table of integrals might have formulas such as this one.
answered Sep 22, 2012 at 7:20
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The magic words are "Lauricella Hypergeometric Function".
answered Sep 22, 2012 at 2:27