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Such Large Proof?

PROVE THAT

\(\displaystyle\int_{0}^{\pi}\)\(\ln\)\((a\pm(b.cos(x)\)))\(dx\)=\(-\)\(\pi\)\(\ln\)\(\dfrac{a+\sqrt{a^{2}-b^{2}}}{2}\)

For \(a\)\(\ge\)b

Note by Parth Lohomi
2 years, 10 months ago

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It seems that I have got a different answer:

Let \[\displaystyle f(z)=\int_0^{\pi} \ln (a\pm z\cos x) dx \] , Where we seek \(f(b)\) and we have \(f(0) = \pi ln a \)

\[\displaystyle f'(z) = \int_0^{\pi} \frac{\pm \cos x}{(a\pm z\cos x)} dx \]

\[\displaystyle = \int_0^{\pi} \frac{\frac{a}{z} \pm \cos x -\frac{a}{z}}{(a\pm z\cos x)} dx = \int_0^{\pi} \frac{dx}{z} -\frac{a}{z} \int_0^{\pi} \frac{dx}{a\pm z\cos x} \]

\[\displaystyle =\frac{\pi}{z} -\frac{a}{z} \int_0^{\pi} \frac{dx}{a\pm z\cos x} \]

Now working on the integral alone:

\[\displaystyle \int_0^{\pi} \frac{dx}{a\pm z\cos x} =2 \int_0^{\frac{\pi}{2}} \frac{dx}{a\pm z\cos 2x} {\color{blue}{(x\to 2x)}} \]

\[\displaystyle = 2 \int_0^{\frac{\pi}{2}} \frac{dx}{a\pm z\cos^2x \mp z\sin^2x} \]

\[\displaystyle = 2 \int_0^{\frac{\pi}{2}} \frac{\sec^2xdx}{a\sec^2x\pm z \mp z\tan^2x} = 2 \int_0^{\infty} \frac{dx}{a(1+x^2)\pm z \mp zx^2} {\color{blue}{(\tan x\to x)}}\]

\[\displaystyle = 2 \int_0^{\infty} \frac{dx}{(a\mp z)x^2 +a\pm z }=\frac{2}{a\pm z} \int_0^{\infty} \frac{dx}{\frac{(a\mp z)}{a\pm z}x^2 +1 } \]

\[\displaystyle \frac{2}{a\pm z} \frac{1}{\sqrt{\frac{(a\mp z)}{a\pm z}}} \int_0^{\infty} \frac{dx}{1+x^2} {\color{blue}{(\sqrt{\frac{(a\mp z)}{a\pm z}} x \to x)}}\]

Simplifying and using \( (a\pm z)(a\mp z) = a^2-z^2 \) , we get:

\[\displaystyle \int_0^{\pi} \frac{dx}{a\pm z\cos x} = \frac{\pi}{\sqrt{a^2-z^2}} \]

Now going back:

\[\displaystyle f'(z) = \frac{\pi}{z} -\frac{a}{z} \int_0^{\pi} \frac{dx}{a\pm z\cos x} = \frac{\pi}{z} -\frac{a}{z} \frac{\pi}{\sqrt{a^2-z^2}} \]

Now integrating back:

\[\displaystyle f(b) = f(b) -f(0) +f(0) = \int_0^b f'(z) dz + \pi \ln a = \int_0^b \frac{\pi dz}{z} -\int_0^b \frac{a\pi dz}{z\sqrt{a^2-z^2}} +\pi \ln a\]

\[\displaystyle = \pi \ln b - \pi \lim_{x\to 0} \ln x +\pi \ln a -\lim_{x\to 0}\int_x^b \frac{a\pi dz}{z\sqrt{a^2-z^2}} \]

Working the integral alone:

\[\displaystyle \int_x^b \frac{a\pi dz}{z\sqrt{a^2-z^2}} = \pi \int_{\arcsin \frac{x}{a}}^{\arcsin \frac{b}{a}} \csc z dz {\color{blue}{(z \to a\sin z )}} \]

Now evaluating this integral will give :

\[\displaystyle = \pi \ln( \frac{\sqrt{a^2-x^2} +a }{x}) - \pi \ln( \frac{\sqrt{a^2-b^2} +a }{b})\]

Now ,the last calculations :

\[\displaystyle f(b) = \pi \ln b - \pi \lim_{x\to 0} \ln x +\pi \ln a -\lim_{x\to 0} \int_x^b \frac{a\pi dz}{z\sqrt{a^2-z^2}} \]

\[\displaystyle = \pi \ln b - {\color{red}{\pi \lim_{x\to 0} \ln x +\pi \ln a }} -{\color{red}{\pi \lim_{x\to 0} \ln( \frac{\sqrt{a^2-x^2} +a }{x})}} + \pi \ln( \frac{\sqrt{a^2-b^2} +a }{b}) \]

\[\displaystyle =\pi \ln b -{\color{red}{\pi \lim_{x\to 0} \ln (\frac{x}{a} \frac{\sqrt{a^2-x^2} +a }{x}) }}+\pi \ln(\sqrt{a^2-b^2} +a) - \pi \ln b \]

\[\displaystyle = -{\color{red}{\pi \ln 2}} +\pi \ln(\sqrt{a^2-b^2} +a) \]

\[\displaystyle =\boxed{\pi \ln (\frac{\sqrt{a^2-b^2} +a}{2}) } \]

Hasan Kassim - 2 years, 9 months ago

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@hasan kassim @Parth Lohomi Could you please tell me the final formula? Parth's and Hasan's results differ by a minus sign.

Ishan Dasgupta Samarendra - 2 years, 4 months ago

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\[\displaystyle =\boxed{+\pi \ln (\frac{\sqrt{a^2-b^2} +a}{2}) } \]

Hasan Kassim - 2 years, 4 months ago

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@Hasan Kassim And this would be for both \(\pm b\cos(x)\)? In other words, \(\displaystyle\int_{0}^{\pi}\)\(\ln\)\((a\pm(b.cos(x)\)))\(dx\)\(=\displaystyle =\boxed{+\pi \ln (\frac{\sqrt{a^2-b^2} +a}{2}) }\)?

Ishan Dasgupta Samarendra - 2 years, 4 months ago

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@Ishan Dasgupta Samarendra yes exactly !

Hasan Kassim - 2 years, 4 months ago

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Help !!@Sandeep Bhardwaj Calvin Lin Ronak Agarwal Sean Ty Krishna Ar Pratik Shastri Pi Han Goh Pranav Arora Mursalin Habib @hasan kassim

Parth Lohomi - 2 years, 10 months ago

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Hey @Parth Lohomi ! , I think I had a mistake, check out now the correct answer, I fixed it :)

Hasan Kassim - 2 years, 9 months ago

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