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

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

I can prove this without much difficulty: $$\int_{0}^{\omega}Re^{-R\cos\theta}d\theta \le \int_{0}^{\omega}Re^{-R\cos\omega}d\theta =\omega Re^{-R\cos\omega} \rightarrow 0 \; (as \quad R \rightarrow \infty)$$ 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

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

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

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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.

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  • $\begingroup$ thank you for your answer. by dominated convergence theorem, you mean that in the real analysis ?can you provide more information about that theorem? $\endgroup$ – 王李远 May 24 '18 at 9:35
  • $\begingroup$ See here: en.wikipedia.org/wiki/Dominated_convergence_theorem. $\endgroup$ – Kelenner May 24 '18 at 9:38
  • $\begingroup$ i have learned that before. but i never know that the theorem can be applied in this situation. $\endgroup$ – 王李远 May 24 '18 at 9:41
  • $\begingroup$ For $K(R)$, we have $R\sin(\theta)^2\exp(-R\sin(\theta))\leq M$ for all $\theta$. is there any problem here?? $\endgroup$ – 王李远 May 24 '18 at 9:47
  • $\begingroup$ Sorry, I do not understand your question. Here the dominant function is the function equal to $M$ for all $\theta$. $\endgroup$ – Kelenner May 24 '18 at 9:55
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This is a classic integral in any textbook. $$\int_{0}^{\frac{\pi}{2}}Re^{-R\sin\theta}d\theta\le\int_{0}^{\frac{\pi}{2}}Re^{-R\frac{2\theta}{\pi}}d\theta=\frac{1}{2}\pi(1-e^{-R})\rightarrow\frac{\pi}{2}\; (as \quad R \rightarrow \infty).$$

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  • $\begingroup$ Yes , but i want to know the exact value. Can you do that? $\endgroup$ – 王李远 May 24 '18 at 9:02
  • $\begingroup$ The trick is $\sin{\theta}\ge\frac{\theta}{\pi/2}$ over $[0,\pi/2]$. $\endgroup$ – MathArt May 24 '18 at 9:03
  • $\begingroup$ Can you prove that? $\endgroup$ – 王李远 May 24 '18 at 9:23
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$$I_R=\int_{0}^{\frac{\pi}{2}}R\,e^{-R\sin(\theta)}\,d\theta=\frac{\pi R}{2} \,(I_0(R)-\pmb{L}_0(R))$$ 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 $1$.

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  • $\begingroup$ thanks for the data, which means the limit should be 1.can you prove this $\endgroup$ – 王李远 May 24 '18 at 9:36
  • $\begingroup$ @王李远. Have a look to my edit. $\endgroup$ – Claude Leibovici May 24 '18 at 10:29
  • $\begingroup$ @王李远. USing the values $10 \leq R \leq 100$, a quick and dirty regression gives $\approx 1+\frac {1.41}{R^{2.10}}$ where the coefficients are highly significant. $\endgroup$ – Claude Leibovici May 24 '18 at 10:49
  • $\begingroup$ that's great ! thanks very much $\endgroup$ – 王李远 May 24 '18 at 11:30
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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}

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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*}

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  • $\begingroup$ That's a great method! thanks! $\endgroup$ – 王李远 May 26 '18 at 4:39

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