Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For an odd function $f$ the integral $\int_{-R}^R f dx$ vanishes if $f$ is continuous on $[-R,+R]$. Now, let $f(x)=\frac{\cos(x)}{x}$ which is odd but discontinuous at 0. How can I show (not) that $\int_{-R}^R f dx = 0$?

share|improve this question
1  
The integral is improper. To compute it, you need to compute $\lim_{t\to 0^-}\int_{R}^tf(x)\,dx$ and $\lim_{t\to 0^+}\int _t^Rf(x)\,dx$. Only if both limits exist will the improper Riemann integral exist and equal their sum. However, I think that $\int_0^1\frac{\cos x}{x}\,dx$ does not converge. –  Arturo Magidin May 22 '12 at 18:54
add comment

3 Answers 3

up vote 2 down vote accepted

The issue is that $f$ is not merely discontinuous at $0$ but unbounded near $0$. Thus what you have is an improper integral. In order for it to converge, by definition, you need $\lim_{a \rightarrow 0^+} \int_a^R f$ to exist.

But this limit does not exist. Hint: for $x$ close to zero, $\frac{ \cos x}{x}$ is close to $\frac{1}{x}$.

Added Of course you also need $\lim_{a \rightarrow 0^-} \int_{-R}^a f$ to exist. But on the one hand, the oddness tells you that if the first limit exists then so will the second. On the other hand, the first limit does not exist, which is enough to show the divergence.)

share|improve this answer
add comment

Your function is unbounded near $0$, in fact $f(x) \approx 1/x$ for small $x$. (By a Maclaurin expansion of $\cos x$.) Hence, your improper integral diverges.

share|improve this answer
add comment

For $-\frac{\pi}{4}\le x\le\frac{\pi}{4}$, we have $\left|\frac{\cos(x)}{x}\right|\ge\frac{1}{\sqrt{2}|x|}$. Therefore, as $\epsilon\to0$, $$ \begin{align} \int_\epsilon^{\pi/4}\frac{\cos(x)}{x}\mathrm{d}x &\ge\int_\epsilon^{\pi/4}\frac{1}{\sqrt{2}\,x}\mathrm{d}x\\ &=\frac{1}{\sqrt{2}}\log\left(\frac{\pi}{4\epsilon}\right)\\ &\to\infty\tag{1} \end{align} $$ and $$ \begin{align} \int_{-\pi/4}^{-\epsilon}\frac{\cos(x)}{x}\mathrm{d}x &\le\int_{-\pi/4}^{-\epsilon}\frac{1}{\sqrt{2}\,x}\mathrm{d}x\\ &=-\frac{1}{\sqrt{2}}\log\left(\frac{\pi}{4\epsilon}\right)\\ &\to-\infty\tag{2} \end{align} $$ Thus, $$ \int_{-R}^{R}\frac{\cos(x)}{x}\mathrm{d}x\tag{3} $$ diverges.

Cauchy Principal Value

Note that the $$ \int_{-R}^{-\epsilon}\frac{\cos(x)}{x}\mathrm{d}x+\int_{\epsilon}^{R}\frac{\cos(x)}{x}\mathrm{d}x=0\tag{4} $$ as $\epsilon\to0$. Equation $(4)$ says that the Cauchy Principal Value of $$ \int_{-R}^{R}\frac{\cos(x)}{x}\mathrm{d}x=0\tag{5} $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.