Integral asymptotic: $\int_0^{\pi} e^{A \sin{x}} dx$ as $A$ goes to inftinity

Question

The exponent $A \sin{x}$ has a single maximum on $[0, \pi]$, so it's natural to use the following method: isolate the extremum point on a small interval $(\pi/2 - \delta_A, \pi/2 + \delta_A)$, express the asymptotic value there and just prove that the remaining integral has smaller order of magnitude (for some suitable $\delta_A$). I believe it should be called saddle-point method or something like that.

The first step didn't cause me too much trouble: using the Taylor expansion of the exponent around $\pi/2$ I was able to conclude that $$ \int\limits_{\pi/2 - \delta_A}^{\pi/2 + \delta_A} e^{A \sin{x}} dx = (1 + o(1)) \sqrt{2\pi} \frac{e^A}{\sqrt{A}}, $$ provided $A \delta_A^3 \underset{A \to \infty}\longrightarrow 0$ and $A \delta_A^2 \underset{A \to \infty}\longrightarrow \infty$.

However, I had a hard time proving that remaining integral, namely, $2 \cdot \int_0^{\pi/2 - \delta_A} e^{A \sin{x}} dx$ has a smaller order of magnitude. I tried using an obvious estimate $A\sin{x} \leqslant Ax$ but it has proven itself too rough. Expanding couple more terms didn't get me anywhere either: integrals involved become just too complicated to even start to estimate and/or evaluate.

Answer 1

Note that $$ \int_0^{\pi /2 - \delta _A } {{\rm e}^{A\sin x} {\rm d}x} \le \frac{\pi }{2}{\rm e}^{A\sin (\pi /2 - \delta _A )} = \frac{\pi }{2}{\rm e}^A {\rm e}^{A(\cos (\delta _A ) - 1)} . $$ By Taylor approximation and your assumptions, we can assert that $$ A(\cos (\delta _A ) - 1) = - \frac{1}{2}\delta _A^2 A + \mathcal{O}(\delta _A^3 A ) \to - \infty $$ as $A\to+\infty$. If, in addition, $$ \delta _A^2 A \gg \log A $$ then $$ {\rm e}^{A(\cos (\delta _A ) - 1)} = o\!\left( {\frac{1}{{\sqrt A }}} \right). $$ For instance, $$ \delta _A = \frac{{\log A}}{{\sqrt A }} $$ is an appropriate choice that satisfies all the requirements.

Answer 2

Let me illustrate a quick trick for replicating Laplace's method. The idea is to expand the exponent about its maximum and rescale it so the integrand converges to the gaussian density. In OP's case, this idea can be implemented as follows:

\begin{align*} \int_{0}^{\pi} \exp(A\sin x) \, \mathrm{d}x &= \int_{-\pi/2}^{\pi/2} \exp(A\cos x) \, \mathrm{d}x \\ &= \int_{-\pi/2}^{\pi/2} \exp(A-2A\sin^2(x/2)) \, \mathrm{d}x \\ &= \frac{e^A}{\sqrt{A}} \int_{-\pi\sqrt{A}/2}^{\pi\sqrt{A}/2} \exp(-2A\sin^2(t/2\sqrt{A})) \, \mathrm{d}t. \tag{$x = t/\sqrt{A}$} \end{align*}

Now by noting that $\left| \sin x \right| \geq c \left| x \right|$ for $|x| \leq \frac{\pi}{4}$ for some constant $c > 0$, the integrand is dominated by the function $e^{-c^2t^2/2}$, which is integrable on all of $\mathbb{R}$. Hence by the dominated convergence theorem,

$$ \int_{-\pi\sqrt{A}/2}^{\pi\sqrt{A}/2} \exp(-2A\sin^2(t/2\sqrt{A})) \, \mathrm{d}t \to \int_{-\infty}^{\infty} \exp(-t^2/2) \, \mathrm{d}t = \sqrt{2\pi}. $$

This proves that the integral is asymptotically

$$ (1 + o(1) \sqrt{\frac{2\pi}{A}}e^A. $$

Answer 3

Too long for a comment and not in the spirit of the question.

From tables of integrals

$$I=\int_0^{\pi} e^{A \sin(x)}\, dx=\pi \, \Big(I_0(A)+\pmb{L}_0(A)\Big)$$ where appear Bessel and Struve functions.

Expanded as series $$I= \sqrt{2\pi}\, \frac{e^A}{\sqrt{A}}\, \sum_{n=0}^\infty \frac{a_n}{2^{b_n} }\, A^{-n}-\frac 2A\, \sum_{n=0}^\infty c_n\, A^{-2n}$$

In $OEIS$

• coefficients $a_n$ form sequence $A274657$ (with a shift of $1$)
• coefficients $b_n$ form sequence $A120738$
• coefficients $c_n$ form sequence $A001818$

A truncated version was presented at the Gordon Research Conference which took place in Tilton (NH) in year $1971$ (no $OEIS$ at that time).