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In the course of work, I deduced such an expression.

$$ p(t)=\int\limits _{-\infty}^{\infty}\begin{cases} 1/x_{max}, & 0\leq t<x\\ 0, & other \end{cases}\cdot dx, $$ where $x \in [0,x_{max}]$ ($x_{max} = 10$ for example),

$t$ - is variable.

I have built a graph that is calculated numerically using python.

import numpy as np
import matplotlib.pyplot as plt

x_max = 10
t = np.linspace(-x_max * 0.1, x_max * 1.1, 1000)

num_point = 1000
pre_p = np.zeros((num_point, t.size))
x = np.linspace(0, x_max, num_point)

for i, this_x in enumerate(x):
    pre_p[i] = np.where((0 <= t) & (t < this_x),  1/x_max, 0)

p = np.trapz(pre_p, x=x, axis=0)


plt.plot(t, p)

enter image description here

I intuitively figured out how this function should look in symbolic form.

$$ p(t)=\begin{cases} 1-\frac{t}{x_{max}}, & 0\leq t<x_{max}\\ 0, & other \end{cases} $$

But how can the integral be taken symbolically according to the rules of mathematics? The thing is, my function is piecewise. And the variable by which I integrate is right in the condition. It would be great to have links to literature or an article, thanks.

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  • $\begingroup$ Your notation of the map being integrated in the first integral is rather confusing. It will help if: (1) you define what $x_{max}, t_{max}$ are and (2) write the function $f(x,t)$ being integrated outside of the integral. Is it a function of two variables $t,x$? $\endgroup$ Apr 20 '21 at 12:47
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    $\begingroup$ Seeing as the graph has no additional area when x=0, it can be done that $Area=\int_0^{10}1-\frac{x}{10} dx$ after finding the line with the points (0,1) and (10,0) as the endpoints to get Area=$(x-\frac{x^2}{20})|_0^{10}$=$10-\frac{10^2}{20}-0$=5 $\endgroup$ Apr 20 '21 at 12:48
  • $\begingroup$ corrected the question. I am interested in taking the integral. $\endgroup$ Apr 20 '21 at 13:00
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Define $$f(t,x) = \begin{cases} \frac 1{x_{\max}},& t\in[0,x]\\0,& t\notin [0,x]\end{cases}$$

Then $$p(t) = \int_0^{x_\max} f(t, x)\,dx$$ Where the limits are from $0$ to $x_\max$, not $-\infty$ to $\infty$, because you said yourself that $x\in [0, x_\max]$. Note that in your definition of $p(t), x$ is a dummy variable, not an actual part of the definition. Its only purpose is to make the notation work. You could switch to a different variable (other that $t$ and $x_\max$, which already have roles) without changing the meaning at all. So in saying $x\in [0, x_\max]$, you are just admitting that you goofed up the limits of the integration.

Now break it out into two cases:

  • $t < 0$ or $t > x_\max$. Then $t\notin [0,x]$ for all $x$ in the limits, so $f(t,x) = 0$ and $\int_0^{x_\max} 0\,dx = 0$
  • $t \in [0, x_\max]$. Then $$\begin{align}p(t) &= \int_0^{x_\max} f(t, x)\,dx \\&= \int_0^t f(t, x)\,dx + \int_t^{x_\max} f(t, x)\,dx\\&=\int_0^t 0\,dx + \int_t^{x_\max} \dfrac 1{x_\max}\,dx\\&= 0 + \dfrac 1{x_\max}(x_\max -t)\\&=1 - \dfrac t{x_\max}\end{align}$$

Putting it together: $$p(t) = \begin{cases}0,& t < 0\\1 - \dfrac t{x_\max}, & 0 \le t \le x_\max\\0,& x_\max < t\end{cases}$$

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  • $\begingroup$ It's great. I began to forget how to take integrals. It's great that there are people who can help. $\endgroup$ Apr 21 '21 at 5:56

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