2

This is probably a silly question, or maybe I am missing a very simple slick trick, but I am trying to see how the following integral is bounded in terms of $\delta$: \begin{equation} \int_{\pi/2+\delta}^{3\pi/2-\delta} x^{R \cos \varphi} d \varphi. \end{equation} $R >0$ is just an arbitrary real number. (This is all to show that a contour integration vanishes). I have tried substitution to no avail as well as absolute value considerations. Any ideas?

EDIT: The only info about $x$ is that $x > 1$. (I think this is supposed to be a straightforward exercise in complex integration but I am a bit rusty, my apologies. The integral is \begin{equation} \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{x^s}{s} ds.) \end{equation}

Numbersandsoon
  • 2,910
  • What is $x$? Is it real or complex, what is its absolute value? – Daniel Fischer Nov 10 '14 at 21:48
  • 1
    I suspect it's about a variant of Jordan's lemma, and you need more than you've stated, namely that the integral multiplied with $R$ remains bounded as $R\to\infty$. – Daniel Fischer Nov 10 '14 at 22:07
  • Edited! Sorry for not including that. – Numbersandsoon Nov 10 '14 at 22:09
  • Yup, a variant of Jordan's lemma. For $x > 1$, you need to close the contour in the left half-plane (and for $c > 0$ that means it encloses the pole at $0$), then use the residue theorem. – Daniel Fischer Nov 10 '14 at 22:29
  • Up to a rotation, it's this. – Daniel Fischer Nov 10 '14 at 22:34
  • @DanielFischer I have done what you suggested, and the above integral in the question is what I get :) Let $S_R$ be the semicircle closing the contour. The integral is $\frac{1}{2\pi i} \int_{c-iR}^{c+iR} \frac{x^s}{s} ds + \frac{1}{2\pi i} \int_{S_R} \frac{x^s}{s} ds=1$. To show that the second integral vanishes, I show that it is less than or equal to $(1/2\pi) \int_{\pi/2}^{3\pi/2} x^{R \cos \varphi} d\varphi$, split it up into three as in $\int_{\pi/2}^{\pi /2 + \delta} + \int_{\pi/2 + \delta}^{3\pi/2 - \delta}+ \int_{3\pi/2 - \delta}^{3\pi /2}$. The first and last are bounded by $\delta$. – Numbersandsoon Nov 11 '14 at 09:16
  • But I was unsure about how to bound the middle one by $\delta$. But I see from your previous answer to the other question linked to above that it might not be necessary to bound it like this. It is possible to obtain a bound with $1/R$? – Numbersandsoon Nov 11 '14 at 09:17
  • 1
    Here, $f(s) = \frac{1}{s}$, so with the notation of the linked question, we have $\mu(R) = \frac{1}{R}$ which gives an overall bound of the form $\frac{C}{R}$. – Daniel Fischer Nov 11 '14 at 12:17

2 Answers2

1

Let $a=|\cos(\pi/2+\delta)|>0$. Then $$ \frac\pi2+\delta\le \varphi\le \frac{3\,\pi}2-\delta\implies-1\le\cos\varphi\le-a $$ and $$ 0<\int_{\pi/2+\delta}^{3\pi/2-\delta}x^{R\cos\varphi}\,d\varphi\le(\pi-2\,\delta)x^{-aR}. $$ Since $x>1$, the last expression converges to $0$ as $R\to\infty$.

Julián Aguirre
  • 76,354
0

If it is the real case...

Because zeros of $\cos x$ are cut off, the function under integral is bounded, the interval is bounded, so...

Przemysław Scherwentke
  • 13,668
  • 5
  • 35
  • 56