Bringing infinite limit inside integral I want to find $$\lim_{R \to \infty} \int_{0}^{\pi} e^{-R\sin (t)}dt.$$ Since $\sin(t)$ is nonnegative on $[0,\pi]$ the integrand vanishes as $R \rightarrow \infty$. So I want to bring the limit under the integral to conclude that the limit is zero, but I'm not sure if it's justified here. I'm not particularly well-versed in this kind of problem so I would appreciate any relevant results/theorems that would help here too. Thanks!
 A: As mentioned in the comments, the dominated convergence theorem is probably the fastest way to justify the interchange of limit and integral. In this particular case that is not necessary, however, as there is an easier method:
You can use the inequality $\sin(x) \geq \frac{2}{\pi} x $ for  $x \in [0,\pi/2]$ (which holds since the sine function is concave on this interval) to find
$$ \int \limits_0^\pi \mathrm{e}^{-R \sin(x)} \, \mathrm{d} x = 2 \int \limits_0^{\pi/2} \mathrm{e}^{-R \sin(x)} \, \mathrm{d} x \leq 2 \int \limits_0^{\pi/2} \mathrm{e}^{-R \frac{2}{\pi} x} \, \mathrm{d} x \, .$$
The latter integral can be computed explicitly and will lead to the desired limit. 
A: Here is an elementary solution that does not involve the dominated convergence theorem but instead uses the fact if $f_n \to f$ uniformly on $[a,b]$, then
$\int_{a}^{b} f_n \to \int_{a}^{b} f$.
To show that 
$$
\lim_{R \to \infty} \int_{0}^{\pi} e^{-R \sin(t)} \, dt = 0,
$$
we will show that for all $\varepsilon > 0$, there exists $R_0$ so that if $R \geq R_0$, then 
$$
\int_{0}^{\pi} e^{-R \sin(t)} \, dt  \leq \varepsilon.
$$
Note: the family of functions $\{ e^{-R \sin(t)} \}_{R > 0}$ converges uniformly to $0$ on the interval $[\varepsilon/4, \pi - \varepsilon/4]$, so 
$$
\lim_{R \to \infty} \int_{\varepsilon/4}^{\pi - \varepsilon/4} e^{-R \sin(t)} \,dt = 0.
$$
That is, there is some $R_0$ so that if $R \geq R_0$, then
$$
\int_{\varepsilon/4}^{\pi - \varepsilon/4} e^{-R \sin(t)} \, dt  \leq \frac{\varepsilon}{2}.
$$
For $R \geq R_0$, we have
$$
\int_{0}^{\pi} e^{-R \sin(t)} \, dt = \int_0^{\varepsilon/4} e^{-R \sin(t)} \, dt
+ \int_{\varepsilon/4}^{\pi - \varepsilon/4} e^{-R \sin(t)} \, dt + \int_{\pi -\varepsilon/4}^{\pi} e^{-R \sin(t)} \, dt  \leq \int_0^{\varepsilon/4} 1 \, dt + \frac{\varepsilon}{2} + \int_{\pi -\varepsilon/4}^{\pi} e^{-R \sin(t)} \, dt= \varepsilon.
$$
(Here, we are using $1$ as an upper bound for the integrand on the intervals $[0, \varepsilon/4]$ and  $[\pi - \varepsilon/4, \pi]$.)
A: To find this limit directly, we divide and conquer with a multi-part estimate; we split the interval into a narrow piece where the integral is small because the interval is small, and a wide piece where the integral is small because the function is small.
First off, I'd rather not deal with both endpoints where $\sin$ is zero, so let's fold it over with symmetry:
$$I(R) = \int_0^{\pi} e^{-R\sin t}\,dt = 2\int_0^{\pi/2}e^{-R\sin t}\,dt$$
Now, we'll split that $\int_0^{\pi/2}$ into $\int_0^{\epsilon/4}+\int_{\epsilon/4}^{\pi/2}$:
\begin{align*}I(R) &= 2\int_0^{\epsilon/4}e^{-R\sin t}\,dt+2\int_{\epsilon/4}^{\pi/2}e^{-R\sin t}\,dt\\
I(R) &\le 2\int_0^{\epsilon/4}1\,dt + 2\int_{\epsilon/4}^{\pi/2}\exp\left(-R\sin \frac{\epsilon}{4}\right)\,dt = \frac{\epsilon}{2}+\left(\pi-\frac{\epsilon}{2}\right)\exp\left(-R\sin \frac{\epsilon}{4}\right)\end{align*}
In both cases, we estimated the function by simply saying that it's less than the largest value on that subinterval.
So now, we need to estimate that exponential term. We have $\sin\frac{\epsilon}{4}\approx \frac{\epsilon}{4}$ - but if we turn that into an inequality, it's a $<$, and that points in the wrong direction for what we need. So instead, estimate $\sin$ by the secant line: $\sin x \ge \frac{2}{\pi}x$ for $0\le x\le \frac{\pi}{2}$. This leads to $\exp\left(-R\sin \frac{\epsilon}{4}\right)\le \exp\left(-R\cdot\frac{2}{\pi}\cdot\frac{\epsilon}{4}\right)$. Then of course $\pi-\frac{\epsilon}{2}\le \pi$, and we can make this term less than $\frac{\epsilon}{2}$ simply by choosing $R$ large enough. For $R\ge \frac{2\pi}{\epsilon}\ln(2\pi)$,
$$I(R) \le 2\int_0^{\epsilon/4}1\,dt + 2\int_{\epsilon/4}^{\pi/2}\exp\left(-R\sin \frac{\epsilon}{4}\right)\,dt \le \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$
As $\epsilon>0$ was arbitrary, that's the definition of convergence to zero, and we're done.
The dominated convergence theorem is of course the easy way; we just use the pointwise convergence to $0$ on the interior (and $1$ at the endpoints) and the fact that all of the functions are less than the integrable function $1$. Then again, this style of argument is worth knowing - it's how we prove the dominated convergence theorem, after all. Running through a few examples like this isn't a bad idea.
Why $\frac{\epsilon}{4}$? Well, you'll probably start out by writing $\epsilon$ - and then when you get an estimate of $4\epsilon$ for the whole thing, go back and divide them all by $4$. The final form of the argument doesn't have to match the scratch version.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
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\begin{align}
\lim_{R \to \infty}\int_{0}^{\pi}\expo{-R\sin\pars{t}}\,\dd t & =
\lim_{R \to \infty}\int_{-\pi/2}^{\pi/2}\expo{-R\cos\pars{t}}\,\dd t =
2\lim_{R \to \infty}\int_{0}^{\pi/2}\expo{-R\cos\pars{t}}\,\dd t
\\[5mm] & =
2\lim_{R \to \infty}\int_{0}^{\pi/2}\expo{-R\sin\pars{t}}\,\dd t
\\[5mm] & =
2\lim_{R \to \infty}\int_{0}^{\infty}\expo{-Rt}\,\dd t
\qquad\pars{~Laplace's\ Method~}
\\[5mm] & =
2\lim_{R \to \infty}{1 \over R} = \bbx{0}
\end{align}

Laplace's Method.

Indeed,
$$
\int_{0}^{\pi}\expo{-R\sin\pars{t}}\,\dd t =
{1 \over 2}\,\pi\bracks{\mrm{I}_{0}\pars{R} - \mrm{L}_{0}\pars{R}}
$$

where $\ds{\mrm{I}_{\nu}}$ and $\ds{\mrm{L}_{\nu}}$ are Bessel and Struve Functions, respectively. 

