6
$\begingroup$

How would one find the zeroes in such cases? I tried some geometric approximations using the graph to no avail.

$$\int_{0}^{2\pi} \ln \left(x + \sin t \right)dt$$

Any help or insight would be appreciated.

$\endgroup$
  • $\begingroup$ Do you mean zeroes of $F(x)=\int_0^x \ln(x+\sin t)dt$? $\endgroup$ – Sonal_sqrt Feb 19 '18 at 3:57
  • $\begingroup$ Not really, the limits are well defined and there is only one variable term which is "x" $\endgroup$ – User 1300135 Feb 19 '18 at 4:00
  • $\begingroup$ For one thing $x\ge 1$ for the integrand to have real values in range $[0, 2\pi]$. $\endgroup$ – Sonal_sqrt Feb 19 '18 at 4:19
  • $\begingroup$ I agree, I'm sure the root lies in between 1 and 2 $\endgroup$ – User 1300135 Feb 19 '18 at 4:31
  • $\begingroup$ @User1300135 I wrote an answer, I am not sure if it is a function in term of x or t. I supposed its a function in term of x $\endgroup$ – shere Feb 19 '18 at 4:46
13
$\begingroup$

Consider the integral as a function of $x$ and then differentiate with respect to $x$. This yields $$f'(x)=\int_0^{2\pi} \frac{1}{x+\sin t}\,dt$$ which can be easily solved by a tangent half angle substitution to yield $$f'(x)=\frac{2\pi}{\sqrt{x^2-1}}.$$ Integrating with respect to $x$ yields $f(x)=2\pi\cosh^{-1}(x)+C.$

To solve for $C,$ consider $f(1).$ The integral is $$\int_0^{2\pi}\ln (1+\sin x)\,dx=\int_0^{2\pi}\ln (1+\cos x)\,dx=2\pi\ln(2)+\int_0^{2\pi}\ln (\cos^2 (x/2))\,dx\\=2\pi\ln(2)+\int_0^{2\pi}\ln (\sin^2 (x/2))\,dx=2\pi\ln(2)+4\int_0^{\pi}\ln (\sin (x))\,dx,$$ which upon combining with the famous result $\int_0^{\pi}\ln(\sin(x))=-\pi\ln(2),$ and $f(1)=C,$ yields $C=-2\pi\ln2$. And lo and behold, $\ln 2= \cosh^{-1}(5/4)!!$ This immediately yields $x=\frac54.$

$\endgroup$
  • $\begingroup$ no the bound of integral is not related to x, your derivative is wrong $\endgroup$ – shere Feb 19 '18 at 5:24
  • $\begingroup$ @shere I am finding the function as a function of $x$, and the first integral is indeed done with respect to $t$. $\endgroup$ – Teoc Feb 19 '18 at 5:24
1
$\begingroup$

For $1 < x < 2$, the integral seems to be $$ -\pi\, \left( 2\,\ln \left( 2 \right) -3\,\ln \left( x+\sqrt {{x}^{2 }-1} \right) -\ln \left( x-\sqrt {{x}^{2}-1} \right) \right) $$ and this is $0$ at $x = 5/4$.

$\endgroup$
  • $\begingroup$ How do you compute this integral? Can you give an outline of the technique? $\endgroup$ – Sonal_sqrt Feb 19 '18 at 4:49
  • $\begingroup$ This seems right, how did you solve it? I've been searching for weeks but got nothing. $\endgroup$ – User 1300135 Feb 19 '18 at 5:07

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.