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I have played around with the differences between definite and indefinite integrals and noted the following:

$$\int C_{0}+C_{1}x\,{\rm d}x-\int_{0}^{x}C_{0}+C_{1}t\,{\rm d}t=0$$

$$\int\ln(x)\,{\rm d}x-\int_{0}^{x}\ln(t)\,{\rm d}t=0$$

$$\int\exp(x)\,{\rm d}x-\int_{0}^{x}\exp(t)\,{\rm d}t=1$$

$$\int x\,\exp(x)\,{\rm d}x-\int_{0}^{x}t\,\exp(t)\,{\rm d}t=-1$$

$$\int\sinh(C\, x)\,{\rm d}x-\int_{0}^{x}\sinh(C\, t)\,{\rm d}t=1/C$$

$$\int\sinh^{-1}(x)\,{\rm d}x-\int_{0}^{x}\sinh^{-1}(t)\,{\rm d}t=-1$$

$$\int\cosh^{-1}(x)\,{\rm d}x-\int_{0}^{x}\cosh^{-1}(t)\,{\rm d}t=\mbox{-}{\mathtt{i}}$$

$$\int\sinh^{-1}(C/x)\,{\rm d}x-\int_{0}^{x}\sinh^{-1}(C/t)\,{\rm d}t=C\,\ln(C)$$

and I was wondering if there is anything special with functions whose result is $0$ or not zero. Actually I started from the last one in a numerical method I was developing, and was intrigued by this extra $C\,\ln(C)$ term that I needed when I calculated the integral with definite limits.

For a polynomial it seems the answer is always zero (I have no proof of it), but for other functions it is not. This is interesting since I thought any continuous function can be approximated as a polynomial using a Taylor's series expansion.

So my question is which functions have their indefinite integral equals the definite one, and which not. Does this have to do with so called transcendental functions.

** EDIT: ** I am using a CAS (computer algebra system) to do the integrations. So maybe the results are implementation specific.

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How do you define $\int f(x) dx$? –  lhf May 25 '11 at 14:43
    
I don't quite understand. For example, when you say that $\int e^x - \int_0^x e^t = 1$, I get lost as the indefinite integral may have any constant attached to it that it wants. I also don't associate only a single definite integral with a function, so that's confusing to me also. –  mixedmath May 25 '11 at 14:44
    
You should also check if all of your definite integrals actually make sense. For example in the second line. –  t.b. May 25 '11 at 14:48
    
@Theo: Good catch! –  mixedmath May 25 '11 at 14:51
    
$\ln(x)$ is integrable near $0$ –  GEdgar May 25 '11 at 16:32

1 Answer 1

up vote 16 down vote accepted

$\int f(x) dx$ doesn't denote an antiderivative, it denotes ALL antiderivatives. If you pick any function which is both integrable and has antiderivatives on an interval containing zero, then

$$\int f(x) dx =\int_0^x f(x)dx + {\mathcal C} \,.$$

so your "difference" should always be ${\mathcal C}$, you forgot the constant in the left integral....

Edit: The right answer to your question is actually all functions have $\int f(x) dx \neq \int_0^x f(x)dx $ because the two terms have different meanings. FTC relates them, but they are different objects.

ADDED after the CAS edit: Since you are using some software, it probably uses a particular antiderivative $F$ to calculate the integrals. Unfortunatelly it wrongly identifies $\int f(x) dx =F$, the actual answer should be $\int f(x) dx =F+ {\mathcal C} $

With any particular choice, the "difference" you get on the right is always $F(0)$. So the functions you are looking for are the ones for which the antiderivative in that particular software has the property $F(0)=0$.

The reason why this is a wrong answer is that while $e^x$ is the choice antiderivative for $e^x$, also $e^x-1$ is an antiderivative of $e^x$ and it would lead to the answer 0. This is exactly the reason why we need that ${\mathcal C} $.

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