# Definitely Integral

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• April 29th 2008, 12:02 PM
Isomorphism
Definitely Integral
To all Integral lovers,

In the Spirit of Integration....here is a popular integral (Thinking)

$\int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}$

Well Enjoy,
Isomorphism

P.S: Elegant methods score can score a lot of 'thank you's :D
• April 29th 2008, 12:19 PM
ThePerfectHacker
• April 29th 2008, 12:32 PM
Isomorphism
Quote:

Originally Posted by ThePerfectHacker

I will wait for some different way of doing it. I have a solution that looks more elegant... but it is almost the same as you did.... (Smirk)
• April 30th 2008, 01:03 AM
Isomorphism
OK i will post my solution since nobody else replied

$I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}$

$I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan \left(\frac{\pi}4 - x\right)} \right)dx}$

$I =\int\limits_0^{\frac {\pi }{4}} \ln \frac2{1 + \tan x} dx$

$I = \int\limits_0^{\frac {\pi }{4}}\ln 2 \,dx - I$

So $I = \frac{\pi}8 \ln2$

Part 2:
$\int^1_0\frac{x^4(1-x)^4}{1+x^2} dx$

This is a funny integral... not hard... but the answer is interesting
People who solve it can comment on the answer rather than showing the answer (Wink)
• April 30th 2008, 02:53 AM
Mathstud28
Quote:

Originally Posted by Isomorphism
OK i will post my solution since nobody else replied

$I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan x} \right)dx}$

$I = \int\limits_0^{\frac {\pi }{4}} {\ln \left( {1 + \tan \left(\frac{\pi}4 - x\right)} \right)dx}$

$I =\int\limits_0^{\frac {\pi }{4}} \ln \frac2{1 + \tan x} dx$

$I = \int\limits_0^{\frac {\pi }{4}}\ln 2 \,dx - I$

So $I = \frac{\pi}8 \ln2$

Part 2:
$\int^1_0\frac{x^4(1-x)^4}{1+x^2} dx$

This is a funny integral... not hard... but the answer is interesting
People who solve it can comment on the answer rather than showing the answer (Wink)

Haha...but how close is that approximation to the real thing...is it less or more making this positive or negative haha
• April 30th 2008, 04:18 AM
Isomorphism
Quote:

Originally Posted by Mathstud28
Haha...but how close is that approximation to the real thing...is it less or more making this positive or negative haha

Hmm actually this is used to approximate 'it' by the use of sandwiching the integral between two series :P
• May 1st 2008, 12:19 PM
Isomorphism
I thought of converting this thread into an "integral collection of my life". So I will be posting integrals here that seem nice to me. Anyone who thinks (s)he has a nice solution, can of course post them here... Thank you :D

The next one in the set:

Part 3:
$\int_0^\infty \frac{dx}{1+x^4}$
• May 1st 2008, 12:27 PM
ThePerfectHacker
Quote:

Originally Posted by Isomorphism
Part 3:
$\int_0^\infty \frac{dx}{1+x^4}$

If I remember correctly: $\int_0^{\infty} \frac{dx}{x^n+1} = \frac{(\pi/n)}{\sin (\pi/n)}$ where $n\geq 2$.
• May 1st 2008, 12:38 PM
Isomorphism
Quote:

Originally Posted by ThePerfectHacker
If I remember correctly: $\int_0^{\infty} \frac{dx}{x^n+1} = \frac{(\pi/n)}{\sin (\pi/n)}$ where $n\geq 2$.

Can you prove that without complex analysis(that is without using residues)?
• May 1st 2008, 01:22 PM
Mathstud28
Quote:

Originally Posted by Isomorphism
Can you prove that without complex analysis(that is without using residues)?

Do you mean by decomposition on the bottom or do you mean by using something else?
• May 1st 2008, 01:31 PM
Isomorphism
$\int_0^\infty \frac{dx}{1+x^4}$ can be solved without complex numbers. I used only pure substitution and algebraic tricks :)

P.S: Unfortunately what I did only holds for this problem (Crying). It does not work for TPH's generalization.
• May 1st 2008, 01:42 PM
Krizalid
Quote:

Originally Posted by Isomorphism

Can you prove that without complex analysis(that is without using residues)?

Combine Gamma & Beta functions plus the Euler Reflection Formula.
• May 2nd 2008, 03:17 PM
Jhevon
Quote:

Originally Posted by Krizalid
Combine Gamma & Beta functions plus the Euler Reflection Formula.

huh? (Speechless)
• May 2nd 2008, 06:05 PM
Krizalid
Let $\int_{0}^{\infty }{\frac{1}{1+x^{n}}\,dx}.$ Make the substitution $z=x^n,$ hence

\begin{aligned}
\int_{0}^{\infty }{\frac{1}{1+x^{n}}\,dx}&=\frac{1}{n}\int_{0}^{\in fty }{\frac{z^{1/n-1}}{1+z}\,dz}\\
&=\frac{1}{n}\beta \left(\frac{1}{n},1-\frac{1}{n}\right)\\
&=\frac{1}{n}\cdot \frac{\Gamma \left( n^{-1}\right)\Gamma\left( 1-n^{-1} \right)}{\Gamma (1)}\\
\end{aligned}

Note aside: I've heard that the Reflection Formula can be only proved via complex analysis. It's a dream to see a solution with elementary tools.
• May 8th 2008, 11:55 PM
Isomorphism
Here is the latest update of "Definitely Integral" (Cool)

Compute I:

$\boxed{I = \int_0^{\pi/4}\ln(\sin x) \cdot \ln(\cos x)\,dx}$

P.S:I think its quite hard :(
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