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plotting - Why does evaluating an integral within a Plot expression take so long?



I want to evaluate


Plot[Integrate[Sin[((1 + 1/2) x)] (1/(2 Sin[(x/2)]) - 1/x), {x, 0, t}], {t, 0, 5}]

but it takes ages to load on Corei3 computer with Mathematica 9.


I'm still a novice when it comes to using Mathematica. Is there a way to get this expression to be evaluated faster? (I'm not used to waiting more than a minute while computing, that's why it feels odd to me).



Answer



Oska nailed it in the comments before me


The trick here is to evaluate the integral only once. In your code it is being evaluated once for every point. You can evaluate the following



Integrate[Sin[((1 + 1/2) x)] (1/(2 Sin[(x/2)]) - 1/x), {x, 0, t}]

The outcome of this is


ConditionalExpression[1/2 (t + 2 Sin[t] - 2 SinIntegral[(3 t)/2]), 
t \[Element] Reals]

You know that t will always be a real number (at least in your plot), so you can just copy paste the first argument of this ConditionalExpression and make a function out of it.


func[t_] := 1/2 (t + 2 Sin[t] - 2 SinIntegral[(3 t)/2])

We can then do



Plot[func[t], {t, 0, 5}]

which gives


enter image description here


To define the function "automatically", you could use the code in Oska's answer. I slightly prefer the following


func[t_] := Evaluate@
Integrate[Sin[((1 + 1/2) x)] (1/(2 Sin[(x/2)]) - 1/x), {x, 0, t},
Assumptions -> t \[Element] Reals]

Which gets rid of the ConditionalExpression and uses SetDelayed rather than Set. The code only works if t does not have value, so watch out.



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