Integrating Trigonometric Substitution and Simplifying Tricky Integrals

[island-boy]

I'm having difficulty with this, trigonometic substitution won't work, neither would integration by parts...
$$\int_{0}^{\infty} \frac{y^2}{1+y^4} dy$$

ETA:
doing trigonometric substitution with $$y^2 = tan\theta$$, I would get
$$\frac{1}{2} \int_{- \frac{\pi}{2}}^{\frac{\pi}{2}} \sqrt{tan \theta} d\theta$$

 

[siddharth]

Don't make the trig sub.

Instead, divide numerator and denominator by $y^2$.
You'll have
$$\int \frac{dy}{y^2 + 1/y^2}$$

There's a tricky step (more like manipulation) to solve this. Here's a hint:
$$y^2 + 1/y^2 = (y+1/y)^2 - 2$$
$$y^2 + 1/y^2 = (y-1/y)^2 + 2$$

Can you play around for a while and take it from here?

 

[island-boy]

hey, thanks for the help siddharth...yeah, i'll try to play around this form and fina a solution.

thanks again

 

[dextercioby]

Contour integration is the solution.

Daniel.

 

[dextercioby]

Also, u might try for a simple fraction expansion.

$$y^{4}+1 =\left(y^{2}+\sqrt{2}y+1\right)\left(y^{2}-\sqrt{2}y+1\right)$$

The result is $\frac{1}{4}\pi \sqrt{2}$

Daniel.

 

[siddharth]

Contour integration is the solution.

Daniel.

You don't need contour integration to solve this.

The trick in integrating
$$\int \frac{dy}{y^2 + 1/y^2}$$

is to write it as

$$(1/2) \int \left( \frac{1-1/y^2}{(y+1/y)^2 - 2} + \frac{1+1/y^2}{(y-1/y)^2 + 2} \right) dy$$

This is very easy to integrate.

 

[benorin]

This is very nice :)

You don't need contour integration to solve this.

The trick in integrating
$$\int \frac{dy}{y^2 + 1/y^2}$$

is to write it as

$$(1/2) \int \left( \frac{1-1/y^2}{(y+1/y)^2 - 2} + \frac{1+1/y^2}{(y-1/y)^2 + 2} \right) dy$$

This is very easy to integrate.