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.
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$\begingroup$ Do you mean zeroes of $F(x)=\int_0^x \ln(x+\sin t)dt$? $\endgroup$– Sonal_sqrtCommented Feb 19, 2018 at 3:57
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$\begingroup$ Not really, the limits are well defined and there is only one variable term which is "x" $\endgroup$– User 1300135Commented Feb 19, 2018 at 4:00
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$\begingroup$ For one thing $x\ge 1$ for the integrand to have real values in range $[0, 2\pi]$. $\endgroup$– Sonal_sqrtCommented Feb 19, 2018 at 4:19
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$\begingroup$ I agree, I'm sure the root lies in between 1 and 2 $\endgroup$– User 1300135Commented Feb 19, 2018 at 4:31
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$\begingroup$ @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 $\endgroup$– shereCommented Feb 19, 2018 at 4:46
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2 Answers
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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 $x=\frac54.$
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$\begingroup$ no the bound of integral is not related to x, your derivative is wrong $\endgroup$– shereCommented Feb 19, 2018 at 5:24
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$\begingroup$ @shere I am finding the function as a function of $x$, and the first integral is indeed done with respect to $t$. $\endgroup$– TeocCommented Feb 19, 2018 at 5:24
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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 $x = 5/4$.
answered Feb 19, 2018 at 4:39
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$\begingroup$ How do you compute this integral? Can you give an outline of the technique? $\endgroup$ Commented Feb 19, 2018 at 4:49
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$\begingroup$ This seems right, how did you solve it? I've been searching for weeks but got nothing. $\endgroup$ Commented Feb 19, 2018 at 5:07