# Difference in answers to integral given by HP 50g emulator and octave

Why is there difference in the answer of the following integral given by Hp 50g emulator and Octave?

It seems to be particularly hard to numerically integrate that function. Wolfram Alpha gives yet another result:
numeric value of integrate tan(ln(x)/cosh(x)) from 0 to 1 - Wolfram|Alpha (wolframalpha.com)

Iâ€™d guess that any numeric integration on a non-finite integrand is probably bound to have pretty high uncertainties. You can see that by slightly changing the waypoints for the integration and observing how much that affects the result.
E.g.:

``````>> f = @(x) tan (log (x) ./ cosh (x));
>> integral(f, 0, 1, 'Waypoints', [0.009, 0.2])
warning: called from
quadgk at line 348 column 7
integral at line 182 column 11

warning: quadgk: Error tolerance not met.  Estimated error 0.696935
warning: called from
quadgk at line 404 column 5
integral at line 182 column 11

ans = -0.5256
``````
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I didnâ€™t understand how did you arrive at the WayPoints [0.009,0.2]?

The function

is discontinued at x=0. But it is continuous at x= 0.009 and x=0.2.
Would you explain your computations of Waypoints?

I didnâ€™t compute the waypoints. I let the function plot by Wolfram Alpha and just took the (approximate) values of the last two singularities as waypoints that might help finding â€śbetterâ€ť steps for the numeric integration:
plot tan(ln(x)/cosh(x)) from 0 to 1 - Wolfram|Alpha (wolframalpha.com)