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 aa≥\geb

#Calculus #Doubts

Note by Parth Lohomi
6 years, 6 months ago

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1 vote

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Comments

It seems that I have got a different answer:

Let f(z)=∫0πln⁡(a±zcos⁡x)dx\displaystyle f(z)=\int_0^{\pi} \ln (a\pm z\cos x) dx , Where we seek f(b)f(b) and we have f(0)=πlnaf(0) = \pi ln a

f′(z)=∫0π±cos⁡x(a±zcos⁡x)dx\displaystyle f'(z) = \int_0^{\pi} \frac{\pm \cos x}{(a\pm z\cos x)} dx

=∫0πaz±cos⁡x−az(a±zcos⁡x)dx=∫0πdxz−az∫0πdxa±zcos⁡x\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}

=πz−az∫0πdxa±zcos⁡x\displaystyle =\frac{\pi}{z} -\frac{a}{z} \int_0^{\pi} \frac{dx}{a\pm z\cos x}

Now working on the integral alone:

∫0πdxa±zcos⁡x=2∫0π2dxa±zcos⁡2x(x→2x)\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{#3D99F6}{(x\to 2x)}}

=2∫0π2dxa±zcos⁡2x∓zsin⁡2x\displaystyle = 2 \int_0^{\frac{\pi}{2}} \frac{dx}{a\pm z\cos^2x \mp z\sin^2x}

=2∫0π2sec⁡2xdxasec⁡2x±z∓ztan⁡2x=2∫0∞dxa(1+x2)±z∓zx2(tan⁡x→x)\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{#3D99F6}{(\tan x\to x)}}

=2∫0∞dx(a∓z)x2+a±z=2a±z∫0∞dx(a∓z)a±zx2+1\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 }

2a±z1(a∓z)a±z∫0∞dx1+x2((a∓z)a±zx→x)\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{#3D99F6}{(\sqrt{\frac{(a\mp z)}{a\pm z}} x \to x)}}

Simplifying and using (a±z)(a∓z)=a2−z2 (a\pm z)(a\mp z) = a^2-z^2 , we get:

∫0πdxa±zcos⁡x=πa2−z2\displaystyle \int_0^{\pi} \frac{dx}{a\pm z\cos x} = \frac{\pi}{\sqrt{a^2-z^2}}

Now going back:

f′(z)=πz−az∫0πdxa±zcos⁡x=πz−azπa2−z2\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:

f(b)=f(b)−f(0)+f(0)=∫0bf′(z)dz+πln⁡a=∫0bπdzz−∫0baπdzza2−z2+πln⁡a\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

=πln⁡b−πlim⁡x→0ln⁡x+πln⁡a−lim⁡x→0∫xbaπdzza2−z2\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:

∫xbaπdzza2−z2=π∫arcsin⁡xaarcsin⁡bacsc⁡zdz(z→asin⁡z)\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{#3D99F6}{(z \to a\sin z )}}

Now evaluating this integral will give :

=πln⁡(a2−x2+ax)−πln⁡(a2−b2+ab)\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 :

f(b)=πln⁡b−πlim⁡x→0ln⁡x+πln⁡a−lim⁡x→0∫xbaπdzza2−z2\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}}

=πln⁡b−πlim⁡x→0ln⁡x+πln⁡a−πlim⁡x→0ln⁡(a2−x2+ax)+πln⁡(a2−b2+ab)\displaystyle = \pi \ln b - {\color{#D61F06}{\pi \lim_{x\to 0} \ln x +\pi \ln a }} -{\color{#D61F06}{\pi \lim_{x\to 0} \ln( \frac{\sqrt{a^2-x^2} +a }{x})}} + \pi \ln( \frac{\sqrt{a^2-b^2} +a }{b})

=πln⁡b−πlim⁡x→0ln⁡(xaa2−x2+ax)+πln⁡(a2−b2+a)−πln⁡b\displaystyle =\pi \ln b -{\color{#D61F06}{\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

=−πln⁡2+πln⁡(a2−b2+a)\displaystyle = -{\color{#D61F06}{\pi \ln 2}} +\pi \ln(\sqrt{a^2-b^2} +a)

=πln⁡(a2−b2+a2)\displaystyle =\boxed{\pi \ln (\frac{\sqrt{a^2-b^2} +a}{2}) }

Hasan Kassim - 6 years, 5 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.

User 123 - 5 years, 11 months ago

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=+πln⁡(a2−b2+a2)\displaystyle =\boxed{+\pi \ln (\frac{\sqrt{a^2-b^2} +a}{2}) }

Hasan Kassim - 5 years, 11 months ago

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@Hasan Kassim – And this would be for both ±bcos⁡(x)\pm b\cos(x)? In other words, ∫0π\displaystyle\int_{0}^{\pi}ln⁡\ln(a±(b.cos(x)(a\pm(b.cos(x)))dxdx==+πln⁡(a2−b2+a2)=\displaystyle =\boxed{+\pi \ln (\frac{\sqrt{a^2-b^2} +a}{2}) }?

User 123 - 5 years, 11 months ago

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@User 123 – yes exactly !

Hasan Kassim - 5 years, 11 months ago

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

Parth Lohomi - 6 years, 6 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 - 6 years, 5 months ago
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