# Calculation of the integral does not occur [closed]

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.

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

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;

• I suggest you do not answer this kind of question which does not show any work from the OP. Jan 8, 2023 at 10:35
• 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 Jan 8, 2023 at 15:47