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How would you go about integrating this?

$\int_{x=-\infty}^{\infty} |1-|\frac{x}{2}||^2 \mathrm dx$

I'm guessing you cannot expand this in anyway.

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    Split the integral into positive and negative parts. Anyway, the integral diverges (for $x>4$ the integrand function is larger than $1$). – 2017-02-21
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    But i still don't know how to do the actual computation. I don't know what to do about the fact that it's squared – 2017-02-21
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    @jdhokia Note that $|x|^2 = x^2$ for all real numbers x. – 2017-02-21
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    $$\left|1-\left|\frac{x}{2}\right|\right|^2=\frac{\left(\left|x\right|-2\right)^2}{4}$$ – 2017-02-21
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    with that^ i think i can do it; but i don't know how you got that :/ – 2017-02-21
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    The integral of such non-negative, continuous and unbounded function over $\mathbb{R}$ cannot be anything else than $+\infty$. – 2017-02-21

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Since $\vert X\vert^2=X^2$ for all $X\in \mathbb R$, your integral becomes:

$$\int_{-\infty}^{+\infty}\left(1-\left\vert \frac x2\right\vert \right)^2\mathrm d x.$$

Then you can split it into two parts since for $x<0$, $\vert x\vert=-x$:

$$\int_{0}^{+\infty}\left(1-\frac x2\right)^2\mathrm d x+\int_{-\infty}^{0}\left(1+\frac x2\right)^2\mathrm d x$$

and both parts diverges to $+\infty$.

At last,

$$\int_{-\infty}^{+\infty}\left\vert1-\left\vert \frac x2\right\vert \right\vert^2\mathrm d x=+\infty.$$

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    I think you were meant to put (1+x/2)^2 on the second integral but yeah i see what you mean :) thanks! – 2017-02-21