How to evaluate the limit $\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta \quad (\text{as } R \rightarrow \infty)$

Question

While doing a mathematical exercise(stein Complex Analysis chapter2,exercise 3), I managed to reduce the problem to the following one:

$$\int_{0}^{\omega}Re^{-R\cos\theta}d\theta \rightarrow 0 \quad (\text{as } R \rightarrow \infty)$$ where $0\le \omega <\frac{\pi}{2}$.

I can prove this without much difficulty: $$\int_{0}^{\omega}Re^{-R\cos\theta}d\theta \le \int_{0}^{\omega}Re^{-R\cos\omega}d\theta =\omega Re^{-R\cos\omega} \rightarrow 0 \quad (\text{as } R \rightarrow \infty)$$ It is crucial that $\omega $ is strictly less than $\frac{\pi}{2}$. This lead me to raise another interesting problem: what the limit will be if we replace $\omega$ by $\frac{\pi}{2}$. After changing $\cos\theta$ to $\sin\theta$ (this doesn't matter), now my question is

$$\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta \rightarrow ? \quad (\text{as } R \rightarrow \infty)$$

I have no idea how to calculate, I even don't know if the limit exists.

Answer 1

Put $I(R)$ your integral and $J(R)=\int_{0}^{\pi/2}R\cos(\theta)^2\exp(-R\sin(\theta))d\theta$, $K(R)=\int_{0}^{\pi/2}R\sin(\theta)^2\exp(-R\sin(\theta))d\theta$. We have $I(R)=J(R)+K(R)$; Note that the function $u\exp(-u)$ is positive and bounded on $[0,+\infty[$, say by $M$.

a) For $K(R)$, we have $R\sin(\theta)^2\exp(-R\sin(\theta))\leq M$ for all $\theta$, and this function goes to $0$ everywhere if $R\to +\infty$. By the Dominated convergence theorem, $K(R)\to 0$ as $R\to +\infty$.

b) For $J(R)$, we integrate by parts: $$J(R)=[(\cos(\theta)(-\exp(-R\sin(\theta))]_0^{\pi/2}-\int_0^{\pi/2}\sin(\theta)\exp(-R\sin(\theta))d\theta$$ We have hence $J(R)=1-\int_0^{\pi/2}\sin(\theta)\exp(-R\sin(\theta))d\theta$. Now apply the dominated convergence theorem to $\int_0^{\pi/2}\sin(\theta)\exp(-R\sin(\theta))d\theta$, and you are done.

Answer 2

This is a classic integral in any textbook. $$\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta\le\int_{0}^{\frac{\pi}{2}}Re^{-R\frac{2\theta}{\pi}}d\theta=\frac{1}{2}\pi(1-e^{-R})\rightarrow\frac{\pi}{2} \quad (\text{as } R \rightarrow \infty).$$

Answer 3

$$I_R=\int_{0}^{\frac{\pi}{2}}R\,e^{-R\sin(\theta)}\,d\theta=\frac{\pi R}{2} \,(I_0(R)-\pmb{L}_0(R))$$ where appear Bessel and Struve functions.

The result tends (quite slowly) to $1$ when $R$ becomes larger and larger as shown in the table below $$\left( \begin{array}{cc} R & I_R \\ 10 & 1.01126 \\ 11 & 1.00909 \\ 12 & 1.00750 \\ 13 & 1.00630 \\ 14 & 1.00538 \\ 15 & 1.00465 \\ 16 & 1.00406 \\ 17 & 1.00358 \\ 18 & 1.00318 \\ 19 & 1.00284 \\ 20 & 1.00256 \end{array} \right)$$

Edit

If you look here, at the bottom of the page, you will find the very interesting asymprotic relation $$\pmb{L}_n(x)=I_{-n}(x)-\frac{x^{n-1}}{2^{n-1}\sqrt{\pi }\, \Gamma \left(n+\frac{1}{2}\right)}$$ Use $n=0$ to get the value of $1$.

Answer 4

This is a straightforward application of Laplace's Method. Since $\sin(\theta)$ is strictly increasing on $[0,\pi/2]$ we have that the main contribution of the integral comes from near $\theta$ around zero. So by applying Laplace's method, we get: \begin{align} \int^{\pi/2}_0 e^{-R \sin\theta}\,d\theta = \frac{1}{R} + o\left( \frac 1 R\right) \end{align} as $R\to\infty$ and therefore: \begin{align} \lim_{R\to\infty} \int^{\pi/2}_0 Re^{-R \sin\theta}\,d\theta = 1 \end{align}

Answer 5

Put \begin{equation*} I(R)=\int_{0}^{\pi/2}Re^{-R\sin \theta}\,\mathrm{d}\theta= \int_{0}^{\pi/4}Re^{-R\sin \theta}\,\mathrm{d}\theta+\int_{\pi/4}^{\pi/2}Re^{-R\sin \theta}\,\mathrm{d}\theta . \end{equation*} Integration by parts yields \begin{gather*} \int_{0}^{\pi/4}Re^{-R\sin \theta}\,\mathrm{d}\theta = \left[\dfrac{-1}{\cos\theta}e^{-R\sin \theta}\right]_{0}^{\pi/4}+ \int_{0}^{\pi/4}\dfrac{\sin \theta}{\cos^2\theta}e^{-R\sin \theta}\,\mathrm{d}\theta=1-\sqrt{2}e^{-R/\sqrt{2}}+\\[2ex]\int_{0}^{\pi/4}\dfrac{\sin \theta}{\cos^2\theta}e^{-R\sin \theta}\,\mathrm{d}\theta . \end{gather*} But according to Lebesgue's dominated convergence theorem \begin{equation*} \int_{0}^{\pi/4}\dfrac{\sin \theta}{\cos^2\theta}e^{-R\sin \theta}\,\mathrm{d}\theta \to 0, \quad R\to\infty \end{equation*} and \begin{equation*} \int_{\pi/4}^{\pi/2}Re^{-R\sin \theta}\,\mathrm{d}\theta \to 0, \quad R\to\infty. \end{equation*} Consequently \begin{equation*} \lim_{R\to\infty}I(R)=1. \end{equation*}

Answer 6

Let $F(R)=\int_0^{\pi/2}e^{-R\cos\theta}\mathbb d\theta$. Note that \begin{align*} F(R) &=\int_0^{\pi/2}\sin\theta e^{-R\cos\theta}\mathbb d\theta+G(R)\\ &=\frac{1-e^{-R}}{R}+G(R), \end{align*} where \begin{align*} 0\leq G(R) &=\int_0^{\pi/2}(1-\sin\theta)e^{-R\cos\theta}\mathbb d\theta\\ &=\int_0^{\pi/2}(1-\cos\theta)e^{-R\sin\theta}\mathbb d\theta\\ &=2\int_0^{\pi/2}\sin^2(\theta/2)e^{-R\sin\theta}\mathbb d\theta\\ &\leq\frac{1}{2}\int_0^{\pi/2}\theta^2e^{-2R\theta/\pi}\mathbb d\theta\\ &\leq\frac{1}{2}\int_0^{\infty}\theta^2e^{-2R\theta/\pi}\mathbb d\theta\\ &=\frac{\pi^3}{8R^3} \end{align*} This proves that, for large $\beta$ we have $$F(R)=\frac{1}{R}+O\left(\frac{1}{R^3}\right),\quad R\to\infty.$$ So $$\lim_{R\to\infty} \int^{\pi/2}_0 Re^{-R \sin\theta}\mathbb d\theta=1.$$