How do I improve Numerical Integration Speed?

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Hi,
I have created a function which will loop several thousand times. I have noticed the time taken to perform the integration necessary is too long to produce the results I need. My integration method is as follows:
fun1= @(x) exp(-(abs(x)).^(1.73)).*((cos(eta1*x)));
fun2= @(x) exp(-(abs(x)).^(1.96)).*((cos(eta2*x)));
R1 =integral(fun1,0,inf);
R4 =integral(fun2,0,inf);
Here I state the function required to be integrated with respect to x and then perform the integral. How would I do this more efficiently. Thank you.
Pierce
  3 Comments
Walter Roberson
Walter Roberson on 25 Aug 2017
You can truncate your integration at
solve(exp(-x^1.73)==eps)
which is about 7 1/2.
However, I do not know if this will help much. In my timing tests it made no difference. If eta1 were large then it probably would make a difference.
David Goodmanson
David Goodmanson on 25 Aug 2017
Hello Peirce, some ways around this depend on the size of eta1 and eta2. What is the range of these variables?

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Accepted Answer

Jan
Jan on 25 Aug 2017
Edited: Jan on 26 Aug 2017
eta1 = 17; % Guessed
fun1 = @(x) exp(-(abs(x)).^(1.73)).*((cos(eta1*x)));
tic
R1 = integral(fun1,0,inf);
toc
Elapsed time is 0.003751 seconds.
It is some percent faster to use a function instead of an anonymous function.
This is not that much, and it was measured on an old Core2Duo. Which speed do you need?
You can reduce the tolerance:
R1 = integral(fun1, 0, inf, 'RelTol', 1e-4, 'AbsTol', 1e-6)
This increases the speed by 10%, but of course it reduces the accuracy.
By the way: Compare these two formulations:
fun1 = @(x) exp(-(abs(x)).^(1.73)).*((cos(eta1*x)));
fun1 = @(x) exp(-(x .^ 1.73)) .* cos(eta1*x);
If the integral has the limits [0, Inf] you can omit the abs. While the two pairs of parentheses around the cos() are not useful, I'd prefer an explicit "-(x .^ y)" to emphasize, that the power has a higher precedence than the unary minus.
  1 Comment
Walter Roberson
Walter Roberson on 26 Aug 2017
My tests with tic/toc had it coming in at about 0.001 seconds on my system, but with timeit() had it coming in at about 0.0004 on my system.

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