Prove that the sequence $\{\int_0^{\pi/2} \sin(t^n)\, dt\}$ converges to 0. Prove that the sequence 
$$
\left\{\int_0^{\pi/2} \sin(t^n)\, dt :n\in\mathbb N\right\}
$$
converges to 0.
 A: First split the integral at $t=1$.  Estimating the first half is simple; since $0 \leq \sin x \leq x$ for $0 \leq x \leq 1$ we have
$$
0 \leq \int_0^1 \sin(t^n)\,dt \leq \int_0^1 t^n \,dt.
$$
For the second half, make the change of variables $t^n = (\pi/2)^n u$ to get
$$
\int_1^{\pi/2} \sin(t^n)\,dt = \frac{\pi}{2n} \int_{(2/\pi)^{\Large n}}^1 \sin\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\,du.
$$
Integrating by parts yields
$$
\begin{align}
&\frac{\pi}{2n} \int_{(2/\pi)^{\Large n}}^1 \sin\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\,du \\
&\qquad = \left.- \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-1}\right|_{(2/\pi)^{\Large n}}^1 \\
&\qquad\qquad + \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \\
&\qquad = - \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \cos\left[\left(\frac{\pi}{2}\right)^n\right] + \frac{\cos(1)}{n} \\
&\qquad\qquad + \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \tag{$*$}
\end{align}
$$
Finally we use the fact that $|\cos(x)| \leq 1$ to estimate the remaining integral;
$$
\begin{align}
\left| \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \right| &\leq -\left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 u^{1/n-2}\,du \\
&= \left(\frac{\pi}{2}\right)^{n-1} - 1 \\
&< \left(\frac{\pi}{2}\right)^{n-1}.
\end{align}
$$
It follows that
$$
\left| \frac{\pi}{2n} \left(\frac{2}{\pi}\right)^n \left(\frac{1}{n}-1\right) \int_{(2/\pi)^{\Large n}}^1 \cos\left[\left(\frac{\pi}{2}\right)^n u \right] u^{1/n-2}\, du \right| < \frac{1}{n}.
$$
Each term in $(*)$ tends to zero as $n \to \infty$, so the result follows.
A: $$
\begin{align}
\left|\,\int_0^{\pi/2}\sin(t^n)\,\mathrm{d}t\,\right|
&=\left|\,\int_0^{(\pi/2)^n}\sin(t)\,\mathrm{d}t^{1/n}\,\right|\\
&=\left|\,\frac1n\int_0^{(\pi/2)^n}t^{1/n-1}\,\mathrm{d}(1-\cos(t))\,\right|\\
&=\left|\,\frac{1-\cos((\pi/2)^n)}{n(\pi/2)^{n-1}}+\frac{n-1}{n^2}\int_0^{(\pi/2)^n}\frac{1-\cos(t)}{t^2}t^{1/n}\,\mathrm{d}t\,\right|\\
&\le\frac2{n(\pi/2)^{n-1}}+\frac{n-1}{n^2}\left(\int_0^1\frac12\,\mathrm{d}t+\int_1^\infty2t^{1/n-2}\,\mathrm{d}t\right)\\
&=\frac2{n(\pi/2)^{n-1}}+\frac{n-1}{n^2}\left(\frac12+\frac{2n}{n-1}\right)\\
&\le\frac9{2n}\tag{1}
\end{align}
$$

With a bit of care, it can be shown that
$$
\begin{align}
\lim_{n\to\infty}\int_0^{(\pi/2)^n}\frac{1-\cos(t)}{t^2}t^{1/n}\,\mathrm{d}t
&=\int_0^\infty\frac{1-\cos(t)}{t^2}\,\mathrm{d}t\\
&=\frac\pi2\tag{2}
\end{align}
$$
which says that asymptotically,
$$
\int_0^{\pi/2}\sin(t^n)\,\mathrm{d}t\sim\frac\pi{2n}\tag{3}
$$

When given Integrate[Sin[t^n],{t,0,Pi/2}], Mathematica 8 returns the answer
$$
\frac{i\pi}{4n}\left(\mathrm{E}_{\frac{n-1}{n}}(-i(\pi/2)^n)-\mathrm{E}_{\frac{n-1}{n}}(i(\pi/2)^n)\right)\tag{4}
$$
where
$$
\begin{align}
\mathrm{E}_n(z)
&=\int_1^\infty\frac{e^{-zt}}{t^n}\,\mathrm{d}t\\
&=\frac{\Gamma(1-n)}{z^{1-n}}-\int_0^1\frac{e^{-zt}}{t^n}\,\mathrm{d}t\tag{5}
\end{align}
$$
However,
$$
\begin{align}
&\mathrm{E}_{\frac{n-1}{n}}(-i(\pi/2)^n) - \mathrm{E}_{\frac{n-1}{n}}(i(\pi/2)^n)\\[4pt]
&=\Gamma\left(\frac1n\right)\frac2\pi\left(i^{1/n}-(-i)^{1/n}\right)-2i\int_0^1\sin((\pi/2)^nt)\,t^{1/n-1}\,\mathrm{d}t\\
&=\frac2\pi\Gamma\left(\frac1n\right)\left(e^{\frac{i\pi}{2n}}-e^{-\frac{i\pi}{2n}}\right)-\frac{4i}{\pi}\int_0^{(\pi/2)^n}\sin(t)\,t^{1/n-1}\,\mathrm{d}t\\
&=\frac{4i}\pi\Gamma\left(\frac1n\right)\sin\left(\frac\pi{2n}\right)-\frac{4in}{\pi}\int_0^{\pi/2}\sin(t^n)\,\mathrm{d}t\tag{6}
\end{align}
$$
Therefore,
$$
\begin{align}
&\frac{i\pi}{4n}\left(\mathrm{E}_{\frac{n-1}{n}}(-i(\pi/2)^n)-\mathrm{E}_{\frac{n-1}{n}}(i(\pi/2)^n)\right)\\
&=-\frac1n\Gamma\left(\frac1n\right)\sin\left(\frac\pi{2n}\right)+\int_0^{\pi/2}\sin(t^n)\,\mathrm{d}t\\
&=-\Gamma\left(1+\frac1n\right)\sin\left(\frac\pi{2n}\right)+\int_0^{\pi/2}\sin(t^n)\,\mathrm{d}t\tag{7}
\end{align}
$$
and it appears Mathematica's answer is off by $-\Gamma\left(1+\frac1n\right)\sin\left(\frac\pi{2n}\right)$.
Actually, $(7)$ decays exponentially, which leaves us with the very good approximation
$$
\begin{align}
\int_0^{\pi/2}\sin(t^n)\,\mathrm{d}t
&\sim\Gamma\left(1+\frac1n\right)\sin\left(\frac\pi{2n}\right)\tag{8}
\end{align}
$$
which supports $(3)$ since $\Gamma\left(1+\frac1n\right)=1-\frac{\gamma}{n}+O\left(\frac1{n^2}\right)$ where $\gamma$ is the Euler-Mascheroni Constant.
A: I made an answer much earlier and I was quite happy of it .... until Antonio Vargas reported to me that my new bounds for integration were totally wrong. I just felt stupid and deleted what I wrote.  
Meanwhile, Antonio Vargas provided a nice and elegant answer to which nothing could to be added.   
But, being upset by my stupidity and trying to be forgiven, I continued working this problem and tried to get a closed form for the integral (this at least will be a small contribution to this problem). 
A CAS arrived to the general formula $$\int_0^{\pi/2} \sin(t^n)\, dt=\frac{i \pi  \left(E_{\frac{n-1}{n}}\left(-i \left(\frac{\pi
   }{2}\right)^n\right)-E_{\frac{n-1}{n}}\left(i \left(\frac{\pi
   }{2}\right)^n\right)\right)}{4 n}$$ which is a real (I have not been able to make the formula simpler) in which appears the exponential integral function. The formula is valid for any integer positive value of $n$. As asked, the limit is zero.
