1. ## Integration

$\int_{-\oe }^{\oe}\frac{1+x^2}{1+x^4}dx$

2. If we don't want to use complex analysis we can proceed in 'elementary' way first remembering that...

$1+x^{4} = (x^{2} - \sqrt{2}\cdot x +1)\cdot (x^{2} + \sqrt{2}\cdot x +1)$ (1)

... and that conducts to the identity...

$\frac{1 + x^{2}}{1+ x^{4}} = \frac{1}{2}\cdot \frac{1}{x^{2} - \sqrt{2}\cdot x +1} + \frac{1}{2}\cdot \frac{1}{x^{2} + \sqrt{2}\cdot x +1}$ (2)

... so that is...

$\int_{- \infty}^{\infty} \frac{1+x^{2}}{1+x^{4}}\cdot dx =\int_{0}^{\infty} \frac{dx}{x^{2} - \sqrt{2}\cdot x +1} + \int_{0}^{\infty} \frac{dx}{x^{2} + \sqrt{2}\cdot x +1} =$

$= \sqrt{2} \cdot |\tan^{-1} (\sqrt{2}\cdot x + 1) + \tan^{-1} (\sqrt{2}\cdot x - 1)|_{0}^{\infty} = \pi \cdot \sqrt{2}$ (3)

Kind regards

$\chi$ $\sigma$

3. Oh , a classic integral

$I = \int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx$

$= 2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx$

$= 2 \int_0^{\infty} \frac{1/x^2 + 1}{1/x^2+x^2}~dx$

$= 2 \int_0^{\infty} \frac{1+ 1/x^2 }{ (x-1/x)^2+2 }~dx$

Sub $x-1/x = t \implies (1+ 1/x^2)dx=dt$

$I = 2 \int_{-\infty}^{\infty} \frac{dt}{t^2+2}$

$= \sqrt{2} [ \tan^{-1}(t/\sqrt{2})]_{-\infty}^{\infty}$

$= \sqrt{2} \pi$

4. ## possible alternative?

Originally Posted by roshanhero
$\int_{-\oe }^{\oe}\frac{1+x^2}{1+x^4}dx$

5. Originally Posted by Pulock2009

Of course , anything you like could be a perfect solution !

Sub $x = \tan(\theta) \implies dx = \sec^2(\theta)~d\theta$

The integral becomes :

$I = 2 \int_0^{\frac{\pi}{2} } \frac{ \sec^4(\theta)}{1 + \tan^4(\theta)} ~d\theta$

$= 2 \int_0^{\frac{\pi}{2} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )}$

$= 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )}$

$= 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ (\sin^2(\theta ) + \cos^2(\theta ))^2 - 2\sin^2(\theta)\cos^2(\theta ) }$

$= 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 2 - \sin^2(2\theta)}$

$= 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 1 + \cos^2(2\theta)}$

$= 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^2(2\theta) + 2\cos^2(2\theta)}$

$= 8 \int_0^{\frac{\pi}{4} } \frac{\sec^2(2\theta) d\theta }{ \tan^2(2\theta) + 2 }$

Sub $\tan(2\theta) = t \implies 2\sec^2(2\theta) d\theta = dt$

$I = 4 \int_0^{\infty} \frac{dt}{t^2+2}$

$= 2\sqrt{2} \cdot \frac{\pi}{2} = \sqrt{2} \pi$

6. ## wonderful!!

Originally Posted by simplependulum
Of course , anything you like could be a perfect solution !

Sub $x = \tan(\theta) \implies dx = \sec^2(\theta)~d\theta$

The integral becomes :

$I = 2 \int_0^{\frac{\pi}{2} } \frac{ \sec^4(\theta)}{1 + \tan^4(\theta)} ~d\theta$

$= 2 \int_0^{\frac{\pi}{2} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )}$

$= 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^4(\theta ) + \cos^4(\theta )}$

$= 4 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ (\sin^2(\theta ) + \cos^2(\theta ))^2 - 2\sin^2(\theta)\cos^2(\theta ) }$

$= 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 2 - \sin^2(2\theta)}$

$= 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ 1 + \cos^2(2\theta)}$

$= 8 \int_0^{\frac{\pi}{4} } \frac{ d\theta }{ \sin^2(2\theta) + 2\cos^2(2\theta)}$

$= 8 \int_0^{\frac{\pi}{4} } \frac{\sec^2(2\theta) d\theta }{ \tan^2(2\theta) + 2 }$

Sub $\tan(2\theta) = t \implies 2\sec^2(2\theta) d\theta = dt$

$I = 4 \int_0^{\infty} \frac{dt}{t^2+2}$

$= 2\sqrt{2} \cdot \frac{\pi}{2} = \sqrt{2} \pi$
one more question:how do u typeout the latex so fast???

7. Originally Posted by Pulock2009
one more question:how do u typeout the latex so fast???

Well , most of them are copied from the previous lines , haha .

8. $
\int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx
$

How did you get this from the above one.I am confused..
$
2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx
$

9. I got this answer,tell me where I am wrong
$
\int_{\infty }^{\infty }\frac{1+x^2}{1+x^4}dx
\int_{\infty }^{\infty }\frac{x^2(1+1/x^2)}{x^2(x^2+1/x^2)}dx
$

$
\int_{\infty }^{\infty }\frac{(1+1/x^2)}{(x-1/x)^2+2}
Put (x-1/x)=u
$

$
\int_{\infty }^{\infty }\frac{du}{u^2+2}
\int_{\infty }^{\infty }\frac{du}{u^2+(\sqrt2)^2}
[\frac{1}{\sqrt2} \tan^{-1}(u/\sqrt{2})]_{-\infty}^{\infty}
\frac{\pi}{\sqrt2}
$

10. Originally Posted by roshanhero
$
\int_{-\infty}^{\infty} \frac{1 + x^2 }{1+x^4}~dx
$

How did you get this from the above one.I am confused..
$
2 \int_0^{\infty} \frac{1 + x^2}{1+x^4}~dx
$
Well, that's easy- since the function being integrated is an even function, the integral from -a to a is just twice the integral from 0 to a.

11. Originally Posted by roshanhero
I got this answer,tell me where I am wrong
$
\int_{\infty }^{\infty }\frac{1+x^2}{1+x^4}dx
\int_{\infty }^{\infty }\frac{x^2(1+1/x^2)}{x^2(x^2+1/x^2)}dx
$

$
\int_{\infty }^{\infty }\frac{(1+1/x^2)}{(x-1/x)^2+2}
Put (x-1/x)=u
$

$
\int_{\infty }^{\infty }\frac{du}{u^2+2}
\int_{\infty }^{\infty }\frac{du}{u^2+(\sqrt2)^2}
[\frac{1}{\sqrt2} \tan^{-1}(u/\sqrt{2})]_{-\infty}^{\infty}
\frac{\pi}{\sqrt2}
$

When i was doing the integral , i faced the same problem :

The explanation is ,

the mapping we have taken $x - 1/x = t$ ,

there is a singularity at $x = 0$

$\lim_{x\to 0^-} (x-1/x) = + \infty$ while

$\lim_{x\to 0^+} (x-1/x) = - \infty$

So we must put $\int_{- \infty }^{+ \infty }$

$= \lim_{a\to 0^+}\left( \int_{- \infty }^{0-a} + \int_{0+a}^{+\infty} \right )$

The substitution gives

$= \int_{ -\infty}^{+\infty} + \int_{-\infty}^{+\infty}$

$= 2\int_{ -\infty}^{+\infty}$

But if we do the substitution first , we will obtain $\int_{ -\infty}^{+\infty}$ which is half of the correct one .

12. this just comes from the fact that given $a>0$ and $f$ being an even integrable function on $[-a,a]$ then $\int_{-a}^a f=2\int_0^a f,$ so there's no much here, the justification comes from dealing with improper integrals.