I'm currently operating with the following integral:

$$\int\frac{u'(t)}{(1-u(t))^2} dt$$

But I notice that

$$\frac{d}{dt} \frac{u(t)}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$


$$\frac{d}{dt} \frac{1}{1-u(t)} = \frac{u'(t)}{(1-u(t))^2}$$

It seems that both solutions are possible, but that seems to contradict the uniqueness of Riemann's Integral.

So the questions are:

  1. Which one of them is the correct integral?
  2. If both are correct, why the solution is not unique?
  3. The pole at $u(t)=1$ has something to say?
  • $\begingroup$ @SanchayanDutta I have edited the title to fit the question a little bit better. If you have ideas for further improvements, go ahead and edit the post. $\endgroup$ – Martin Sleziak Sep 7 '15 at 20:13
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    $\begingroup$ "but that seems to contradict the uniqueness of Riemann's Integral" whoever said antiderivatives were unique? $\endgroup$ – PyRulez Sep 8 '15 at 0:12
  • $\begingroup$ @PyRulez ocw.mit.edu/courses/mathematics/… $\endgroup$ – Dargor Sep 8 '15 at 0:14
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    $\begingroup$ @PabloGalindoSalgado Notice that it says "unique up to a constant" $\endgroup$ – PyRulez Sep 8 '15 at 0:17
  • $\begingroup$ @PyRulez Yup, yup. I thought that your question were refered to that result, not the constant. Sorry. $\endgroup$ – Dargor Sep 8 '15 at 0:19

It is not really a contradiction, since difference of the two functions is constant: $$ \frac1{1-u(t)} - \frac{u(t)}{1-u(t)} = \frac{1-u(t)}{1-u(t)}=1. $$ (Derivative of a constant function is zero. Primitive function is determined uniquely up to a constant.)


I have not verified that the derivatives are correct, but notice that $$\frac1{1-u(t)}-\frac{u(t)}{1-u(t)}=1$$

that is, the difference between both antiderivatives is constant. This implies that the derivatives of both functions are the same.


$$\frac{d}{dt} \frac{u(t)}{1-u(t)}=\frac{d}{dt} \left(\frac{1}{1-u(t)}-1\right)=\frac{d}{dt} \frac{1}{1-u(t)}$$

Note that $1$ is just a constant so it vanishes. But when computing the antiderivative you will specify the constant according to initial conditions.


An equivalent way to see this, but from a slightly different perspective:

$$\frac{1}{1 - u(t)} = \frac{1 - u(t) + u(t)}{1 - u(t)} = 1 + \frac{u(t)}{1 - u(t)}$$

Perhaps this technique is worth seeing, namely, tinkering with an expression using additive inverses; specifically, by subtracting then adding back $u(t)$ to the numerator, we come to see that the two expressions differ only by a constant (i.e., $1$).

This technique occasionally presents itself when integrating; for example,

$$\int \frac{x}{x+1} dx = \int \frac{x+1-1}{x+1} dx = \int 1 - \frac{1}{x+1} dx$$

where the leftmost integrand is, in my estimation, a bit less friendly than the rightmost.


The other answers are excellent and clearly explain that indeed anti derivatives of a function may differ by a constant. I wanted to add that your "problem" can be introduced both in a "forward" and a "backward" fashion.

You gave a "backward" example, that is when two clearly not equal functions upon differentiation produce the same function. But working the other way around can also introduce the "problem", that is integration of the same function (and omitting the constant) may still produce a constant difference in the resulting functions. To illustrate what I mean consider the following tempting conclusion

$$\text{Let } f(x) = x, g(x) = x \text{ then } f(x) = g(x) \text{ and } \int f(x)dx = \int g(x)dx$$

We now apply the results of integration theory on both sides of the equation (which should result in applying the same operations on the symbols, right?) and decide not to add an extra constant to either of both. If we let the resulting expressions be denoted respectively by $F(x)$ and $G(x)$, we might be tempted to say that $F(x) = G(x) = \frac{1}{2}x^2$, but this conclusion is also not entirely true because the results depend on the operations applied. Moreover, the applied operations might be (unknowingly) forced upon us by the form of the equation!

To see this, consider the differential equation $y' = y$. Suppose I have the solution $z = 2e^x$, now we might be tempted to say

$$ \frac{z'}{z} = 1 \rightarrow \int \frac{z'}{z}dx = \int dx \xrightarrow{\text{set both integration constants 0}} \log(z) = x \rightarrow z = e^x \rightarrow 2e^x = e^x $$

which is obviously false.

I have sometimes seen students struggling with this, so I thought I might as well add it. Hope it helps! :)


I will explain the concept using the derivative since you are already pretty familiar with that. Lets define to $\int \:f\left(x\right)dx$ to be some fucntion where $\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)$. (Note this is not the exact definition of an anti-derivative, but an intuitive way of thinking about it.)

So if $g\left(x\right)=x^2,\:\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)=2x$ And in reverse,$\int \:f\left(x\right)dx=x^2$

But what happens if $g\left(x\right)=x^2+5,\:\frac{d}{dx}\left(g\left(x\right)\right)=f\left(x\right)=2x$, Notice the 5 is a constant and disappears when taking the derivative. If we apply the reverse, we have no way of getting back the 5. In reverse you still get $\int \:f\left(x\right)dx=x^2$.

So now here comes the problem, when you are going in reverse you have no idea what the constant is. This is purely because the constant disappears when taking the derivative. For example $\frac{d}{dx}\left(x^2+8\right)=\frac{d}{dx}\left(x^2+5\right)=\frac{d}{dx}\left(x^2+1\right)$! So when finding the Anti-Derivative you will get a function plus or minus some constant.


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