Is there a symbol for integrating and setting $C=0$?

Question

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 !

Answer 1

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. :)

Answer 2

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.

Answer 3

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, $\int_0^x \sin t\, dt=1-\cos x$.

2. Using integral from $-\infty$

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

This integral would not converge though for the most functions.

3. 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.

UPDATE.

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.

Answer 4

$C=0$ 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.

Answer 5

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:

$\operatorname{const}+\operatorname{const}=\operatorname{const}$

$\operatorname{const}\cdot\operatorname{const}=\operatorname{const}$

$\operatorname{const}-\operatorname{const}=\operatorname{const}$

$\operatorname{const}/\operatorname{const}=\operatorname{const}$

$\operatorname{const}+a=\operatorname{const}$

$a\operatorname{const}=\operatorname{const}$

if $a\ne0$ but

$0\operatorname{const}=0$

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