Too often it's useful to just set $C$ to $0$ after integrating, so I was wondering if there is a symbol for that?

Formally this would mean a shorter way of writing: $\int_{t_0}^x f(t)\,\mathrm{d}t = F(x), (F(t_0) = 0)$

This is useful when integrating sums of integral and splitting these as in: $\int x^2+2\,\mathrm{d}x$.

$(1)\quad \int x^2+2\,\mathrm{d}x = \color{green}{\int_{t_0}^x t^2\,\mathrm{d}t}+\color{red}{\int2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x + C}$

But this changes the variable from $x$ to $t$, which isn't very nice. Another way is:

$(2)\quad\int x^2+2\,\mathrm{d}x = \color{green}{\int x^2\,\mathrm{d}x}+\color{red}{\int 2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3} + C_1} + \color{red}{2x + C_2} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x} + \color{blue}{C}, \,(\color{green}{C_1}+\color{red}{C_2} = \color{blue}{C})$

Which is sometimes just written as:

$(3)\quad\int x^2+2\,\mathrm{d}x = \color{green}{\int x^2\,\mathrm{d}x}+\color{red}{\int 2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x} + \color{blue}{C}$

But which I find is missing a step, as it doesn't explain where the $\color{blue}{C}$ comes from. If I could just choose an antiderivative from the first integral $\left(\color{green}{\int x^2\,\mathrm{d}x}\right)$ without a constant, this would be fine.

I was thinking maybe

$(4)\quad \int_{t_0}^x f(t)\,\mathrm{d}t = F(x), (F(t_0) = 0) \overset{_\text{def}}{=} \int_0 f(x)\,\mathrm{d}x = F(x)$

(With the property that $\int_0 f(x)\,\mathrm{d}x + C = \int f(x)\,\mathrm{d}x$)

$(5)\quad \int x^2+2\,\mathrm{d}x = \color{green}{\int_0 x^2\,\mathrm{d}x}+\color{red}{\int x^2+2\,\mathrm{d}x} = \color{green}{\dfrac{x^3}{3}} + \color{red}{2x + C}$

Or in general:

$(6)\quad \int f_1(x)+f_2(x)+\cdots+f_n(x)\,\mathrm{d}x = \int_0 f_1(x)\,\mathrm{d}x+\int_0 f_2(x)\,\mathrm{d}x+\cdots\int f_n(x)\,\mathrm{d}x$

So is there perhaps a better way to write this in an unambiguous way, if not: would my own notation be confused anywhere? (In any case it'd be used only for personal use).

Thanks in advance !

  • 1
    $\begingroup$ If you don't want to change variables, I think you can use $\int_0 f(x) dx$, as you suggested, or $\int_{x_0} f(x) dx$ (if $x_0$ such that $F(x_0)=0$ is not obvious from context). It seems that there is no standard notation for this. You just have to define your shortening somewhere. This may conflict with integration over set ($\int_\mathbb R$, $\int_\Omega$) and both variants conflict with each other if you mix them, but I don't see any more problems. $\endgroup$
    – BartekChom
    Dec 28, 2014 at 23:25
  • $\begingroup$ @BartekChom Thanks. Part of me was worried as I've yet to learn about $\int_{\mathbb{R}}$ and such, so I was worried I might get myself confused in the future. $\endgroup$
    – Frank Vel
    Dec 29, 2014 at 0:07

5 Answers 5


There is no such notation for setting $C=0$.

While there are already several good answers, I believe one thing everyone has failed to mention is exactly why there isn't such a notation.

I believe the following example best illustrates the problem:

$$\int \sin(2x)\,dx = \int 2\sin(x)\cos(x)\,dx = \sin^2(x)+C$$ where we used the substitution: $u=\sin(x)$ and so $du=\cos(x)\,dx$.

On the other hand, $$\int \sin(2x)\,dx = \int 2\sin(x)\cos(x)\,dx = -\cos^2(x)+C$$ where we used the substitution: $u=\cos(x)$ and so $du=-\sin(x)\,dx$.

Of course both $\sin^2(x)$ and $-\cos^2(x)$ are antiderivatives of $\sin(2x)$. Notice that they differ by a constant: $\sin^2(x) =-\cos^2(x)+1$.

But with your proposed notation which one do we get when $C=0$? The answer is ambiguous.

To pick out a particular antiderivative you need to pick a base point like: $F(x) = \int_0^x f(t)\,dt$ so that $F'(x)=f(x)$ and $F(0)=0$. This is exactly the choice you have advocated above. So you could do this. If this is what you want, you're notation would be: $F(x)=$ "$\int_0^x f(t)\,dt$". I guess you could abbreviate this as "$\int_0^x f$" and everyone would understand what you mean.

However, when computing indefinite integrals, there is no universally accepted base point. Why? It might seem like $x_0=0$ is a good choice for a base point, but it wouldn't work for cases like $\int 1/x\,dx = \ln|x|+C$.

The ambiguous indefinite integral notation is there because it allows us to be lazy and avoid making a choice. :)

  • $\begingroup$ Yeah but in some instances it is nice to just choose an integral (even if that too is an ambiguous and non-unique choice), and I felt $\int_0^x f(t)\mathrm{d}t$ (as I gave an example of) is somewhat long and is somewhat confusing as you change the variables. $\endgroup$
    – Frank Vel
    Dec 28, 2014 at 23:51
  • $\begingroup$ @fvel: "Just choose an integral" is exactly what the indefinite integral notation means! But that is something different from your "set $C=0$" which sounds like you're expecting to get a definite result from that without making any choice. $\endgroup$ Dec 29, 2014 at 0:11
  • $\begingroup$ @HenningMakholm The indefinite integral is a set of functions, all differing by a constant. I just want a single function from that set, and I want to do that by setting $C$ to be $0$. So the choice is for $C$ to be $0$, even if that yields non-unique answers. $\endgroup$
    – Frank Vel
    Dec 29, 2014 at 1:16
  • $\begingroup$ @fvel: You seem to persist in thinking that "setting $C$ to be $0$" is meaningful and picks out a particular function from the set -- it doesn't. There is no canonical correspondence between members of the set and values of $C$. $\endgroup$ Dec 29, 2014 at 1:21
  • $\begingroup$ @HenningMakholm I'm aware that there wouldn't be a unique choice, but it still be a function, and not a set, which is what I'd want. $\endgroup$
    – Frank Vel
    Dec 29, 2014 at 2:05

If you specifically want an antiderivative with $F(x_0)=y_0$, rather than just any antiderivative, it seems to be easy to write $y_0+\int_{x_0}^x f(t) \, dt$ instead of $\int f(x)\,dx$.

There doesn't seem to be much room for making that significantly more compact (especially when you have $y_0=0$), so the conceptual cost of having a specialized notation for this case would probably not be worthwhile.

For your own private notes you can of course use whichever notation you like.

  • $\begingroup$ Is there no commonly used shorter way of writing $y_0+\int_{x_0}^x f(t) \mathrm{d}t, (y_0 = 0)$? This is sometimes just written just as $\int f(x)\mathrm{d}x$, but this seems like abuse of notation and feels ambiguous... $\endgroup$
    – Frank Vel
    Dec 28, 2014 at 20:57
  • $\begingroup$ @fvel: For $y_0=0$ you don't need to write the "$0+$", of course. So then it's just $\displaystyle \int_{x_0}^x f(t)\,dt$, which can't be much shorter given that there has to be somewhere to write the $x_0$. $\endgroup$ Dec 28, 2014 at 21:03
  • $\begingroup$ I would necessarily also write that $F(x_0) = 0$, is there no common way of shortening this? $\endgroup$
    – Frank Vel
    Dec 28, 2014 at 21:10
  • $\begingroup$ @fvel: I don't get your "necessarily". If you write $\int_{x_0}^xf(t)\,dt$, at $x=x_0$ you automatically get 0 from $\int_{x_0}^{x_0}\cdots\,dt$. No need to write that down separately. $\endgroup$ Dec 28, 2014 at 21:11
  • $\begingroup$ Right, but I still have to change the variable, any way to get around that? $\endgroup$
    – Frank Vel
    Dec 28, 2014 at 21:19

As a result of the procedures of integration you can get just any function from all the functions that are the antiderivatives. If you want to make integration constant always fixed, I suggest the following variants:

  1. Using integral from zero

$$\int_0^x f(t) dt$$

This is quite standard if you do not bother a lot, but the finction may be undefined in 0. This also directly answers your questtion, because you want the integral to be zero in zero.

But it has a disadvantage that for instance, while all derivatives of $e^x$ are equal to the same function, $e^x$, the integral will be $\int_0^x e^t dt=e^x-1$ to meet the requirement. Similarly, $\int_0^x \sin t\, dt=1-\cos x$.

  1. Using integral from $-\infty$

$$\int_{-\infty}^x f(t) dt$$

This integral would not converge though for the most functions.

  1. Using the natural integral

$$\int_N f(x)dx=D^{-1}[f](x)=f^{(-1)}(x)=\frac{i}{2\pi}\int_{-\infty}^{+\infty} \frac{e^{- i \omega x}}{\omega} \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

This fixes the integration constant in the most natural way, but to find such integrals you would need extensive knowledge in Fourier analysis. Additionally the expression may not always converge or require extended technique to take a generalized integal.

Another formula that produces natural integral for a different set of functions is

$$f^{(-1)}(x)=\sum_{m=0}^{\infty} (-1)^m \sum_{k=0}^m\binom mk(-1)^{m-k}f^{(k)}(x)$$ Sometimes you would have to find natural integral for a function with a parameter and then analytically continue it into the area of parameter's values whetre the expressions do not converge.

That said, natural integral will usually produce the simplest function you want, for instance,

$$(\sin x)^{(-1)}=-\cos x$$ $$(\cos x)^{(-1)}=\sin x$$ $$(e^x)^{(-1)}=e^x$$ $$(x^a)^{(-1)}=\frac{x^{a+1}}{a+1}$$

But there can be exceptions. For instance,

$$\left(\frac 1 x\right) ^{(-1)}=\log|x|+\gamma$$

where $\gamma$ is the Euler-Mascheroni constant.


The Fourier formula can be employed using Wolfram Alpha. Follow the link: http://www.wolframalpha.com/input/?i=I+InverseFourierTransform[FourierTransform[sin+t%2C+t%2C+w]%2Fw%2C+w%2C+x]

and replace sin t with any function u want.

  • $\begingroup$ Isnt $\binom{-1}{m}$ just $(-1)^m$? Or how do you define that? $\endgroup$ Dec 28, 2014 at 20:47
  • $\begingroup$ @Henning Makholm yes, it is just a vestige of using the same formula for integrals/derivatives of other orders. $\endgroup$
    – Anixx
    Dec 28, 2014 at 20:52
  • $\begingroup$ x @Anixx: Perhaps you could give a hint about which formula it is you're generalizing here. Something usually used for fractional derivatives? $\endgroup$ Dec 28, 2014 at 20:59
  • 1
    $\begingroup$ @fvel it will work for your condition that a function to be zero in zero. The function $1-\cos x$ is zero in zero, so it satisfies your requirement even though it seems more complicated than just $-\cos x$. If you want the most natural integration constant ever, there is no simpler way than to use the natural integral. $\endgroup$
    – Anixx
    Dec 28, 2014 at 21:15
  • 1
    $\begingroup$ @fvel and what antiderivative you want? Any? Do u consider constant term is zero in $\ln ax$? Note that $\ln ax = \ln x + \ln a$. You can take any function with constant term and make from it a function without a constant term. $\endgroup$
    – Anixx
    Dec 28, 2014 at 21:58

$C=0$ by itself doesn't enforce anything; whence there is no extra notation for it. It is only after you have specified your preferred primitive that $C=0$ makes sense.


I realized that you may be confused by the need to use several different constant terms when summing integrals and the like. Some computer algebra systems solve this problem the following way.

You can define a new mathematical object $\operatorname{const}$ with the following properties:







if $a\ne0$ but


etc. You can think of it like of expression $0/0$ if you like, because any number satisfies equation $0x=0$.

Then when integrating, instead of $+C$, write $+\operatorname{const}$ and remember its properties when making manipulations.

For this purpose for instance, Mathematica has an embeeded object C: http://reference.wolfram.com/language/ref/C.html


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