The integral of $\frac{\cos(x)}{x}$ from $0$ to $1$

So I have a test next week and I saw this question:

proof that $\ \int_{0}^{1} \frac{\cos(x)}{x} \, \mathrm{d}x$ is not convergent.

So first of all I know this is improper integral.

When I saw the solution (a student done few years ago) it says that:

since $\forall\ 0 \leq x \leq 1$ $0 \leq \frac{\cos(x)}{x}$

And therefore we can use the limit comparison in order to check if it is not convergent. So

$\lim \frac{\frac{\cos(x)}{x}}{\frac{1}{x}}= 1$ while $x\rightarrow 0$

and because $\ \int_{0}^{1} \frac{1}{x} \, \mathrm{d}x$ is not convergent then

$\ \int_{0}^{1} \frac{\cos(x)}{x} \, \mathrm{d}x$ is not covergent also.

but as far as I know he should checked that

$\lim \frac{\frac{\cos(x)}{x}}{\frac{1}{x}} < \infty$ while $x\rightarrow \infty$ in order to determine something and not zero.

Am I missing something, because clearly he is right in the final answer but is way is true ?

Thanks in advanced

-
Why would he take $x \to \infty$? Your integral goes from $x=0$ to $x=1$, not to $x = \infty$. – Robert Israel Aug 29 '13 at 15:12

5 Answers

Assume the contrary that $\displaystyle \int_0^1 \frac{\cos x}{x} dx$ does exist. Let $M > 0$ be any upper bound of the integral. Notice $\cos(x) \ge \cos(1) > 0$ on $[0,1]$, we find for any $\epsilon \in (0,1)$, we have

$$M \ge \int_0^1 \frac{\cos x}{x} dx \ge \int_\epsilon^1 \frac{\cos x}{x} dx \ge \int_\epsilon^1 \frac{\cos(1)}{x} dx = \cos(1)\log\frac{1}{\epsilon}\tag{*}$$

If we choose $\epsilon < e^{-\frac{M}{\cos(1)}}$, L.H.S of $(*)$ will be greater than $M$. This is a contradiction and hence the original assumption that $\displaystyle \int_0^1 \frac{\cos x}{x} dx$ exists is wrong.

When you get into latter part of your course which allow you to be a little bit sloppy about the most elementary stuff. You can prove the same statement by first claiming $\displaystyle \frac{\cos x}{x} \ge 0$ on $[0,1]$ and then compare your integral with one that is known to diverge. In this case, $\displaystyle \int_0^1 \frac{\cos(1)}{x} dx\;$ is again a good choice.

-
Thank u very much !! – wantToLearn Aug 29 '13 at 16:43

If $\lim_{x \to 0} (\cos x / x) / (1/x)$ is a non-zero constant (like 1), and both functions have a single discontinuity at $0$, then $\int_0^1 \frac{\cos x}{x} dx$ is indeed divergent if and only if $\int_0^1 \frac{1}{x} dx$ is divergent.

-
Thanks for the help !! – wantToLearn Aug 29 '13 at 16:44

You do not need to check what is happening beyond the region of integral.

Technically you could define a new function $g(x)=\frac{1}{x}$ within the interval, and anything you want, say $1$, outside. Likewise you could do the same with $f(x)=\frac{\cos(x)}{x}$ in $[0,1]$, and $0$ outside.

Then your condition on $x\to \infty$ will hold for $\frac{f(x)}{g(x)}$, which is essentially $\frac{cos(x)/x}{1/x}$ in the interval $[0,1]$.

-
helped me a lot. thanks !! – wantToLearn Aug 29 '13 at 16:44

Well, note that $\cos(x)$ > $\cos(1)$ for your specific interval. Therefore, we can use the fact that $\frac{\cos(x)}{x} \geq \frac{\cos(1)}{x}$. If we can show that the integral of the lesser function, $\frac{\cos(1)}{x}$ is divergent, then we can show that the integral of the greater function must also be divergent. Therefore, we must attempt to show that the following is divergent:

$\int_{0}^{1}\frac{\cos(1)}{x}\mathrm{d}x$. This is fairly simple. As you pointed out, $\int_{0}^{1}\frac{1}{x}\mathrm{d}x$ is divergent, because of the behavior of $\ln(x)$ at $x=0$. We can rewrite $\int_{0}^{1}\frac{\cos(1)}{x}\mathrm{d}x$ by factoring out the $\cos(1)$:

$\cos(1)\int_{0}^{1}\frac{1}{x}\mathrm{d}x$.

The above is clearly divergent, since we know that $\int_{0}^{1}\frac{1}{x}\mathrm{d}x$ is divergent. Thus, we have proven that $\int_{0}^{1}\frac{\cos(1)}{x}\mathrm{d}x$ is divergent. Since

$\frac{\cos(x)}{x} \geq \frac{\cos(1)}{x}$, the integral of the greater function, $\frac{\cos(x)}{x}$ must also be divergent from 0 to 1.

-
Thank u !! I realy appreciate that. – wantToLearn Aug 29 '13 at 16:44

Hint : Since $\cos(x)$ is positive for $\, 0<x<1$

$$\frac{\cos(x)}{x} \leq \frac{1}{x}$$

-
That inequality goes in the wrong direction. You might use $\dfrac{\cos x}{x} \ge \dfrac{\cos 1}{x}$. – Robert Israel Aug 29 '13 at 15:13
To clarify, the inequality is true, but it doesn't help answer the problem. – Mark Lakata Aug 29 '13 at 15:26
Since $cos(x)$ would be less than one, the inequality is true. However, we would want to show that the desired function is greater than the one that diverges. Knowing that it is less than 1/x, the divergent function, is inconclusive. – ra1nmaster Aug 29 '13 at 15:30