# Is this precise indispensable actually independent of a parameter? Just how can it be revealed?

I intend to find a wonderful straightforward expression for the precise indispensable

$$\int_0^\infty \frac{x^2\,dx}{(x^2-a^2)^2 + x^2}$$

Now, I can numerically calculate this indispensable, and also it appears to merge to $\pi/2$ for all actual values of $a$. Is this indispensable in fact constantly equivalent to $\pi/2$? Just how can I show this?

Additionally, why does Wolfram Alpha offer me something that shows up to rely on $a$? Exists an excellent factor it does not remove $a$?

With apologies to Robin Chapman.

The integrand is an also function of $x$, so we can incorporate from $-\infty$ to $\infty$ and also take fifty percent. The integrand often tends to $1/z^2$ for huge $z$ and also as the size of a huge arc is $\pi z$ the payment of the arc often tends to absolutely no. So we simply require to incorporate over the upper fifty percent aircraft. The deposits are remedies to $0=(x^2-a^2)^2+x^2= (x^2-i x -a^2)(x^2+i x-a^2)$. If $x_1$ and also $x_2$ are remedies to the first square, after that at $x_1$ the 2nd polynomial amounts to $2i x_1$, and also the deposit at $x_1$ is $x_1^2/2ix_1(x_2-x_1)=x_1/2i(x_2-x_1)$. The amount of deposits at $x_1$ and also $x_2$ is consequently $1/2i$. Currently we simply keep in mind that both posts in the upper fifty percent aircraft are without a doubt the remedies $x_1$ and also $x_2$ (which are $i (\pm\sqrt{-a^2+1/4} + 1/2)$) ). Therefore the shape indispensable is $\pi$, and also the initial indispensable is $\pi/2$.

Yes it holds true!

Allow

$$\displaystyle I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$

Make the replacement $\displaystyle x = \dfrac{a^2}{t}$

We get

$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^6}{t^4\left(\dfrac{a^4}{t^2} + \left(\dfrac{a^4}{t^2} - a^2\right)^2\right)} \ \text{dt} = \int_{0}^{\infty} \dfrac{a^2}{t^2 + (a^2 - t^2)^2} \ \text{dt}$$

i.e.

$$\displaystyle I = \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2} \ \text{dx}$$

Therefore $$\displaystyle 2I = \int_{0}^{\infty} \dfrac{x^2}{x^2 + (x^2-a^2)^2} \ \text{dx} + \int_{0}^{\infty} \dfrac{a^2}{x^2 + (a^2 - x^2)^2}\ \text{dx}$$

$$ = \int_{0}^{\infty} \dfrac{x^2 + a^2}{x^2 + (x^2-a^2)^2} \ \text{dx}$$

$$\displaystyle = \int_{0}^{\infty} \dfrac{1 + \dfrac{a^2}{x^2}}{1 + \left(x-\dfrac{a^2}{x}\right)^2} \ \text{dx}$$

Making the replacement $\displaystyle t = x - \dfrac{a^2}{x}$

Gives us

$$\displaystyle 2I = \int_{-\infty}^{\infty} \dfrac{\text{dt}}{1 + t^2} = \pi$$

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