# 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
3 years, 6 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

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 :

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

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

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

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

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

- 3 years, 5 months ago

Log in to reply

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

- 2 years, 12 months ago

Log in to reply

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

- 2 years, 12 months ago

Log in to reply

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

- 2 years, 12 months ago

Log in to reply

yes exactly !

- 2 years, 12 months ago

Log in to reply

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

- 3 years, 6 months ago

Log in to reply

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

- 3 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...