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Here we have an elementary limit that Mathematica simply doesn't want to compute it.
What solutions might I have to fix that? Could you help here? As you can easily guess,
the limit is precisely $0$.

Below you may find the code I used

N[Limit[Integrate[Sin[t^n], {t, 0, Pi/2}], n -> Infinity, Assumptions -> n \[Element] Integers]]
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Since you already know the answer, why do you need Mathematica to compute it? Limit and Integrate don't work together, only separately. If the inner Integrate can't finish, Limit can't do anything. –  Szabolcs Feb 22 '14 at 15:25
@Szabolcs Out of curiosity. I expect Mathematica gives the correct answer, don't you think? This is not really true since one can put both Limit and Integrate together and they work, but not always. –  Chris's sis the artist Feb 22 '14 at 15:32
I'm not aware of any examples where Integrate doesn't work alone, Limit doesn't work on the integrand, but Limit[Integrate[...],...] does work. If you are, please let me know. –  Szabolcs Feb 22 '14 at 15:47
I'm more puzzled by the difference between integ = Integrate[Sin[t^n], {t, 0, Pi/2}, Assumptions -> {n \[Element] Integers, n > 0}] and integ2 = 1/n Integrate[y^(1/n - 1) Sin[y], {y, 0, (Pi/2)^n}, Assumptions -> {n \[Element] Integers, n > 0}]. Try Table[integ // N // Chop, {n, 1, 100, 10}] and Table[integ2 // N // Chop, {n, 1, 100, 10}]. –  b.gatessucks Feb 22 '14 at 16:43

1 Answer 1

up vote 4 down vote accepted

"What solutions might I have to fix that?" As Szabolcs says, if theIntegratecommand does not return an expression, other thanIntegrate[...], thenLimithas no chance to operate. So, I would first find the integral by supplyingIntegratewith sufficient assumptions:

Integrate[Sin[t^n], {t,0,Pi/2}, Assumptions :> {n>=1}]

The assumption of integer $n$ is not sufficient, but $n\ge1$ is sufficient. The result involves the exponential integrals

ExpIntegralE[(n-1)/n, +I (Pi/2)^n]
ExpIntegralE[(n-1)/n, -I (Pi/2)^n]

and a Gamma function term

Gamma[1+1/n] Sin[Pi/(2n)]

The limit as $n\rightarrow\infty$ of the Gamma function term is clearly 0, which Mathematica easily finds. The limit of either ExpIntegralE term cannot be found by Mathematica (I'm running v9). However, if you write

Limit[ExpIntegralE[a, I x], x->Infinity]


Limit[FunctionExpand[ExpIntegralE[a, I (Pi/2)^n]], n->Infinity]

Mathematica succeeds in finding the correct result for you.

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