I'm a beginner with Matlab, and I'm trying to solve the following problem.

I'd like to define a function $$F(x) = \int_0^{+\infty} \frac{\sin t}{1 + xt^4} \, dt$$ and plot it on the interval $[1,3]$.

I wrote the following code:

x = 1:0.01:3;


for i = 1:201

xi = 1 + (i-1)/100;

y(i) = quadgk(@(t)sin(t)./(xi*t.^4 + 1),0,inf); 



This looks ridiculously complicated to me, and there's no obvious way to compute values of $F(x)$ after this.

I'd like to know how experienced Matlab users would program this.

  • 1
    $\begingroup$ Matlab's main purpose is not to do symbolic calculations. It can be done, as you've shown. But usually (at least in my experience) Matlab is used for numerics rather than symbolics. $\endgroup$ – NicNic8 Oct 28 '14 at 20:03
  • $\begingroup$ In this case, I'm mainly interested in numerical calculations. I'd like to simplify the syntax, even if internally Matlab is doing the equivalent of a for loop. $\endgroup$ – user188062 Oct 28 '14 at 20:03
  • $\begingroup$ You do not need to integrate up to $\infty$ (I guess you only want a plot and not an superaccurate answer) Since the function falls off like $1/(1+xt^4)$ it should be enough to go up to say $xt^4 \approx 10^2 - 10^3$, i.e. try to integrate from $t=0$ to $t=10$. It should speed up the computation. $\endgroup$ – Winther Oct 28 '14 at 20:05
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    $\begingroup$ Doesn't Matlab do that automatically? I mean, can't it bound the integrand so that it decides where to cut off the integral? $\endgroup$ – user188062 Oct 28 '14 at 20:06
  • $\begingroup$ Matlab's quadgk does allow integration over infinite intervals (as long as the integrand decays rapidly enough). Its methods are rather more sophisticated, I think, than picking a finite cutoff. $\endgroup$ – Robert Israel Oct 28 '14 at 20:24

I might do it this way:

integrand = @(x) (@(t) sin(t) ./ (1+x .*t .^4)); F = @(x) quadgk(integrand(x),0,Inf); x = 1:0.01:3; y = arrayfun(F,x); plot(x,y)

enter image description here

  • $\begingroup$ This doesn't work for me. I get "Error using ^ Inputs must be a scalar and a square matrix. To compute elementwise POWER, use POWER (.^) instead." and some other stuff. I'm using a 2012 version of Matlab. Could that be why? Or is it something to do with declaring the types of the variables? $\endgroup$ – user188062 Oct 28 '14 at 20:28
  • $\begingroup$ Did you have to enter any commands like "syms x", etc., before executing this? $\endgroup$ – user188062 Oct 28 '14 at 20:37
  • $\begingroup$ Oops, left out some "."'s. Fixed now. $\endgroup$ – Robert Israel Oct 28 '14 at 21:10
  • $\begingroup$ It works now. If I understand correctly, integrand is a function which, evaluated at x, gives you a function of t. So integrand(x) is actually a function of t. Is that right? $\endgroup$ – user188062 Oct 28 '14 at 21:21
  • $\begingroup$ Yes, that's right. $\endgroup$ – Robert Israel Oct 28 '14 at 21:23

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