I have thought about the following equalities for a while and maybe you can suggest are they true or false, and help me to solve some inconsistencies which occur. I am making one question for three equalities because they are connected.
So:
1) Is $$\int f(x)\,dx = b + \int f(x)\,dx $$
where $f$ is some function and $b$ is some number.
2) Is $$\int f(x)\,dx - \int f(x)\,dx = 0$$
where $f$ is some function. This equality may seem a bit obvious but in connection with other one it can cause some problems (inconsistencies).
3) I definitely know that:
$$\int \frac{1}{x}\,dx = 1 + \int \frac{1}{x}\,dx$$
and I know this because you can easily prove it using integration by parts.
And now, this equality in combination with (2) leads to the following false equality $$0=1$$
So there can be two posibilities: either (2) is false or I am doing some technical mistakes when I conclude that $$0=1$$ So what is the mistake ?
Thanks in advance
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There are hundreds of such proofs $0=1$ by integration, e.g., here, or here, or here with $1/x$. – Dietrich Burde May 17 '18 at 15:06
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@Dietrich Burde So what is the technical mistake I am doing ? – Юрій Ярош May 17 '18 at 15:11
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Have you ever heard the expression "plus a constant". – fleablood May 17 '18 at 15:23
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"I definitely know that: ∫1xdx=1+∫1xdx and I know this because you can easily prove it using integration by parts." No, you don't. You know and proved $\int \frac 1x dx = 1 + \int \frac 1x dx + C$. "plus a constant". In this case the constant is $-1$. – fleablood May 17 '18 at 15:25
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@fleablood I think that is exactly the point of "equation 2)" which leads to a misunderstanding in addition to the nonspecific notation. – Mundron Schmidt May 17 '18 at 15:26
1 Answers
The problem arises at the notation of $\int f(x)dx$. On the first sight, it seem that you get a unique function $F(x)=\int f(x)~dx$, but that is false. There are infinte many primitives for a function $f$ which are equation up to a constant. If you don't need to specify the constant, then you can use the notation $\int f(x)~dx$. But if you write down $$ \int f(x)dx = b+\int f(x)dx, $$ then you can indeed find primitives on the left and right side such that the equation holds, but then the same term has different meaning in one equation and you can't just say that they has to be the same because of the same notation.
Therefore, you should use a better notation like $$ F(x)=\int_a^x f(t)~dt. $$ Here, $F$ is a specific primitive of $f$ and now $$ F(x)=b+F(x) $$ implies $b=0$.
The same problem apears in 3). The equation is true in the sense that there exists different primitives on the LHS and RHS of the equation such that the equation holds. But further the primitive on the LHS and RHS are not the same and you can't just drop it.
Finally, 2) has the same problem. If you use the same primitive for both integrals, the equation is true.
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+1 To add to the problem, a function may have an antiderivative on several disconnected intervals, and then the definite integral trick will fail. There is a separate constant on each interval where the integrand is continuous). That's a variation on this: a function that is differentiable at every point where it is defined, and whose derivative is zero everywhere, is not necessarily constant (can make a funny exam question). – Jean-Claude Arbaut May 17 '18 at 15:25
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You are right. I thought of a continuous function $f$ defined on an interval $[a,b]$, because from the question, I thought that he has to handle problems first on that level. If the domain is more complex, you have to be even more precise and careful. – Mundron Schmidt May 17 '18 at 15:31
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Thanks for the answer, it clarifies for me all, but it shows me that indefinite integral notation is a bit bad. Am I wrong ? – Юрій Ярош May 17 '18 at 15:38
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No, you are right. The notation isn't good for every situation. Sometimes it is fine if there is no misunderstanding. But you have always to be aware of the context and exact definition of a notation. – Mundron Schmidt May 17 '18 at 15:49