-4
$\begingroup$

It is necessary to calculate the a definite integral of the probability distribution function:

\begin{align} I &= \int\limits_{\xi=-\infty}^{-1} \frac{\mu \, d\xi}{2 \cdot (1+\mu^2 \cdot ((\xi-m) \cdot k)^2)^{3/2}}. \end{align}

syms psii mu m ku
f = mu./(2.*(1+mu.^2.*((psii-m)*ku).^2).^(3/2));
P2 =int(f, psii, -inf, -1)

Calculation of the integral does not occur.

$\endgroup$
4
  • 2
    $\begingroup$ It is so simple ! $\endgroup$ Jan 8, 2023 at 9:19
  • $\begingroup$ If you know, tell me why the calculation is not happening? $\endgroup$
    – Антон
    Jan 8, 2023 at 9:31
  • 2
    $\begingroup$ What is the antiderivative ? How do you want to be helped if you do not show your work. Using Matlab is not doing mathematics. $\endgroup$ Jan 8, 2023 at 9:34
  • $\begingroup$ I can find a primitive one like this: P2 =int(f, psii) $\endgroup$
    – Антон
    Jan 8, 2023 at 10:30

1 Answer 1

0
$\begingroup$

With the integrand $f(\xi)$

$$f(\xi )\text{=}\frac{\mu }{2 \left(\mu ^2 (k (\xi -m))^2+1\right)^{3/2}}$$

we will get the antiderivative

$$F(\xi )=\int f(\xi ) \, d\xi =\frac{\mu (\xi -m)}{2 \sqrt{k^2 \mu ^2 (m-\xi )^2+1}}$$

Finally we treat the limits to calculate the definite integral

$$\int_{-\infty }^{-1} f(\xi ) \, d\xi =F(-1)-\underset{x\to -\infty }{\text{lim}}F(x)=\frac{1}{2} \mu \left(\frac{1}{\sqrt{k^2 \mu ^2}}-\frac{m+1}{\sqrt{k^2 \mu ^2 (m+1)^2+1}}\right)$$

Working MATLAB code:

syms xsi mu m k
f = mu/(2*(1+mu^2*((xsi-m)*k)^2)^(3/2));
F = int(f, xsi);
l1 = limit(F, xsi, -1);
l2 = limit(F, xsi, -inf);
result = l1 - l2;
$\endgroup$
2
  • 1
    $\begingroup$ I suggest you do not answer this kind of question which does not show any work from the OP. $\endgroup$ Jan 8, 2023 at 10:35
  • $\begingroup$ Thank you very much, kind man. I find it strange that matlab does not calculate a certain integral immediately, but calculates if the limits are substituted manually $\endgroup$
    – Антон
    Jan 8, 2023 at 15:47

Not the answer you're looking for? Browse other questions tagged .