# How can I find the constant of integration?

I'm attempting a novel approach to some tough integration problems. I'm using the idea of series expansions to help integrate. In other words, I will attempt to approximate integration by integrating the series expansion of an integrand, rather than direct integration or standard numerical methods.

I believe I can approximate integration of a series very easily, compared to the other methods. However, there's a catch. I will use at least two different series expansions. One for the lower limit of integration, and one for the upper limit. Now, when I attempt to integrate these expansions, the constant of integration comes into play, and it's not obvious what it is. Since I am using at least two different series expansions, the constant of integration may differ for each expansion. So I'm wondering if there is an easy way to get the constants of integration without much more work. Any help, ideas, or suggestions are welcome.

EDIT

A few additional notes... I know ahead of time that the series will converge. I consider that I could integrate in sections, like quadrature, while still using the series to aid in integration. However, I am considering the idea of only using only the endpoints, with two different series. So the constants of integration would be different for each series. If I could somehow find them or find how they differ relative to one another, that would save me the trouble of breaking the integral into sections and using something akin to conventional numerical methods.

• You are calculating a definite integral. You must use the same constant of integration for the lower limit as for the upper limit. Effectively, you need not worry about the constant of integration, any series that has the right derivative will do. But to repeat, you must use the same series in calculating at upper limit, lower limit. – André Nicolas Aug 8 '11 at 15:52
• I think you have in mind to replace the integrand by a series approximation, which is certainly an idea worth pursuing. However I don't think your thinking about it is ripe enough for an answer to be given. The constant of integration really won't matter if you are developing an approximation to a definite integral. If you are approximating an antiderivative, then the consistency of a constant across a boundary where you switch the approximating series can be a valid problem. Basically you would choose the constants so the results agree at the boundary where they switch (continuity). – hardmath Aug 8 '11 at 16:00
• Series expansions are sometimes a very useful thing. However, for most nice functions of any complexity, even getting the first few terms is difficult. And there are issues of convergence. Look for example at $\int_3^{11}\frac{1}{1+x} \,dx$. The usual series for $1/(1+x)$ does not converge for $x$ in our interval. – André Nicolas Aug 8 '11 at 16:08
• @André: you can always use the Laurent series, which is also pretty usual :) – Mariano Suárez-Álvarez Aug 8 '11 at 16:22
• @Mariano Suarez-Alvarez: True, it depends on the meaning of "usual." I was trying to point to difficulties at the OP's likely level in his/her studies. – André Nicolas Aug 8 '11 at 16:40

If you are talking about looking for the anti-derivative (though you also mentioned upper and lower limits), then you just write some constant $C$ after all your calculations. E.g. if $$f(x) = \sum\limits_{k=0}^\infty a_k x^k$$ then $$F(x) = \int f(t)\,dt = \sum\limits_{k=0}^\infty\frac{a_k}{k+1}x^{k+1} +C.$$ The idea of an anti-derivative is the following: $$\int\limits_{a}^bf(t)\,dt = F(b)-F(a)$$ for any anti-derivative $F$ of the function $f$. Since any anti-derivative is determined up to a constant, you just should pick one anti-derivative before calculating $F(b)-F(a)$. In our example, you can put $C=1$ or $C=2$ and the result will stay the same.