0
$\begingroup$

Evaluate the integral $\int_0^1 \cos(\ln(x)) \, dx$

I was able to evaluate the improper integral which is:

$$\frac{x\left(\sin \ln x + \cos \ln x\right)}{2}$$

I was using the substitution $u = \ln x$, and afterward I did integration by parts twice and got the result:

$$\frac{e^u\left( \sin(u) + \cos(u) \right)}{2}$$

Applying $x=1$ we get $u = 0$ and applying $x=0$ we get $u=-\infty$.

So how can it be calculated?

$\endgroup$
  • $\begingroup$ So what exactly do you want to evaluate? Do you wish to simplify the result you've got? $\endgroup$ – Kugelblitz Feb 26 '15 at 13:05
  • $\begingroup$ The value of the integral between $0$ to $1$. $\endgroup$ – AlonAlon Feb 26 '15 at 13:06
  • $\begingroup$ Wolfram Alpha says you get approximately 0.5 wolframalpha.com/input/… I guess it considered the terms for your split improper integral to be 1/2 * 1 - 1/2 * 0 by inputting 1 and 0 respectively and for some reason disregarding the problems which arise when you input zero for the natural log.. $\endgroup$ – Kugelblitz Feb 26 '15 at 13:22
3
$\begingroup$

Set

$$\begin{align} u &= \ln(x) \implies e^{u}du = dx \\ x &= 0 \implies u = -\infty \\ x &= 1 \implies u = 0 \end{align}$$

Hence, you get

$$I = \int_{-\infty}^{0} e^{u}\cos(u) du$$

Integrating by parts twice, first with $v = e^{u}$, $w' = \cos(u)$ and secondly with $v = e^{u}$, $w' = \sin(u)$

$$\begin{align} I &= \int_{-\infty}^{0} e^{u}\cos(u) du \\ &= e^{u}\sin(u) \biggr|_{-\infty}^{0} - \int_{-\infty}^{0} e^{u}\sin(u) du \\ &= 0 - \int_{-\infty}^{0} e^{u}\sin(u) du \\ &= -\bigg[ -e^{u}\cos(u)\biggr|_{-\infty}^{0} + \int_{-\infty}^{0} e^{u}\cos(u) du \biggr] \\ &= 1 - \int_{-\infty}^{0} e^{u}\cos(u) du \\ &= 1 - I \end{align}$$

Therefore

$$\begin{align} I &= 1 - I \\ \implies I &= \frac{1}{2} \\ \end{align}$$

Wolfram gives the same result.

$\endgroup$
  • $\begingroup$ Thank you very much! @Mattos $\endgroup$ – AlonAlon Feb 26 '15 at 13:27
  • $\begingroup$ @AlonAlon Happy to help. $\endgroup$ – mattos Feb 26 '15 at 13:27
1
$\begingroup$

Hint

$$ \lim_{x\to 0^+} -\frac{x}{2}\leq \lim_{x\to 0^+} \frac{x\sin(\ln(x))}{2}\leq \lim_{x\to 0^+} \frac{x}{2} $$

Works also for $x\cos(\ln(x))$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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