Is an integral itself a function or a constant?

I have three cases:

Q1: Say I have $f:\mathbb R \rightarrow \mathbb R$ and $\int_a^bf(t)\, dt$. Should I write $$ g(t)=\int_a^bf(t)\, dt \tag 1 $$ where $g:\mathbb R\rightarrow \mathbb R$. Or should I write $$ a=\int_a^bf(t)\, dt \tag 2 $$ where $a$ is a constant, $a\in \mathbb R$.

Q2: If I have $x:\mathbb R\rightarrow \mathbb R$ and an improper integral $\int_{-\infty}^{\infty}x(t)\, dt$, should I write $$ h(t)=\int_{-\infty}^{\infty}x(t)\, dt \tag 3 $$ where $h:\mathbb R\rightarrow \mathbb R$. Or with a constant $b\in \mathbb R$ $$ b=\int_{-\infty}^{\infty}x(t)\, dt \tag 4 $$

Q3: Also an improper integral of $y:\mathbb R\rightarrow \mathbb R$, so $\int_{-\infty}^{\infty}\lvert y(t)\rvert ^2\, dt$. Is the following correct $$ p(x)=\int_{-\infty}^{\infty}\lvert y(t)\rvert ^2\, dt \tag 5 $$ where $p:\mathbb R\rightarrow \mathbb R$. Or with a constant $c\in\mathbb R$ $$ c=\int_{-\infty}^{\infty}\lvert y(t)\rvert ^2\, dt \tag 6 $$


  • $\begingroup$ None of the integrals in your question are indefinite integrals. The integrals in Q2 and Q3 are improper definite integrals. $\endgroup$ – Rahul Feb 7 '18 at 10:38
  • $\begingroup$ Hi @Rahul Changed it. $\endgroup$ – JDoeDoe Feb 7 '18 at 10:40

Three cases:

We use the integral sign without bounds to mean the 'antiderivative' or the 'Newton integral' of a function, the function whose derivative is equal to the original function. This is a function, we usually use the same name for the argument as the integration parameter. This is sometimes called the indefinite integral, but strictly speaking is not (see below).

$$ g(x) = \int f(x) dx $$

We use the integral sign with bounds (finite or infinite), to mean the definite integral between those bounds. This is a number. As @Rahul points out in a comment, improper definite integrals (those with infinite bounds) are also numbers.

$$ a = \int_{0}^{\infty} f(x) dx $$

We use the integral sign with two bounds, one of which is a variable, to mean the integral between a fixed bound and a variable bound. This defines a function. The function's argument is the variable bound, which can't also be the integration parameter. This is the 'indefinite integral' in the strict sense. This function is probably an antiderivative of the original function, but is not defined to be one. It is a theorem that the antiderivative and the indefinite integral are the same, and holds only under certain conditions.

$$ g(t) = \int_{-1}^{t} f(x) dx $$

  • 1
    $\begingroup$ The kind of explanation everyone needs. +1 $\endgroup$ – Paramanand Singh Feb 7 '18 at 16:29

In $Q_1, Q_2$ and $Q_3$, the integrals are constants.


Assuming $a$ and $b$ are constants, in all three cases you've noted the integral is a constant.

It can be a function, if the function "under" the integral sign depends on $t$ and something else, or if the limits are variable. For example:

  • $\int_a^b f(x,t)dt$ is a function of $x$.
  • $\int_x^y f(t)dt$ is a function of $x$ and $y$.
  • Sometimes you can have a combination, e.g. $\int_{g(x,y)}^{h(y,z)}f(x,y,z,t)dt$ is a function of $x, y, z$.

Crucially, the integral it is never a function of $t$ - $t$ is "bound" by $dt$.


Definite integrals as in your examples are a way to denote a number, exactly like a sum or a limit. In this sense

$$\int _a^b f(t)\,\mathrm dt$$

is a constant. The variable $t$ has no meaning outside of the integral. However, stuff like this exists:

$$g(x):=\int_a^b f(x,t)\,\mathrm dt.$$

Here the integrand depends on $x$ and $t$ but the integral is only over $t$, hence the $x$ exists outside the integral and can be used to define the function $g(x)$.

  • $\begingroup$ Hi! In the integral $g(x):=\int_a^b f(x,t)\,\mathrm dt$, does it explicit mean $x$ is a function of $t$, i.e. $g(x(t)):=\int_a^b f(x(t),t)\,\mathrm dt$? Or is $x$ just a variable (exactly as you wrote)? $\endgroup$ – JDoeDoe Feb 11 '18 at 9:24
  • $\begingroup$ @JDoeDoe $x$ is just a variable. If $x$ would be a function, then $$\int_a^b f(x(t),t)\,\mathrm dt $$ would be a constant. $\endgroup$ – M. Winter Feb 11 '18 at 12:04

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