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Sorry for my bad english.

We consider $f_{\alpha} : R \rightarrow [0,+\infty]$ defined by $f_{\alpha}(x) = \dfrac{\sqrt{|x|}}{1-x} \mathbb{1}_{[0,\alpha]}(x)$ for $\alpha \in [0,1]$.

We want to write the integral $\int_{R} f_{\alpha} d \lambda$ as a sum of series with positives terms.

I really don't know how to do it ! I thought to write the function in the form $$\dfrac{\sqrt{|x|}}{1-x} = \frac{1}{1-x}\times\sqrt{|x|}$$ but without success.

Someone could help me ? Thank you in advance.

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  • $\begingroup$ Isn't t easier to write $\int_{0}^{\alpha}{\,\sqrt{\,x\,}\, \over 1 - x}\,\mathrm{d}x$ with $\alpha \in \left(\,{0,1}\,\right)$ ?. It's quite readable !!!. $\endgroup$ Apr 30, 2017 at 1:37

2 Answers 2

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Hint. One may recall that $$ \sum_{n=0}^\infty x^n=\frac{1}{1-x}, \qquad |x|<1, $$ then one is allowed to write, for $0\le\alpha<1$, $$ \int_{\mathbb{R}}f_{\alpha}(x) d \lambda = \int_{\mathbb{R}}\dfrac{\sqrt{|x|}}{1-x} \mathbb{1}_{[0,\alpha]}(x)\:dx=\sum_{n=0}^\infty \int_{0}^\alpha x^n\sqrt{|x|}dx=\sum_{n=0}^\infty\frac{\alpha^{n+3/2}}{n+3/2}. $$

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  • $\begingroup$ @MélanieDelaCheminée You are welcome. $\endgroup$ Apr 24, 2017 at 13:57
  • $\begingroup$ Toujours aussi efficace, je ne m'en lasse pas :) $\endgroup$
    – ParaH2
    Apr 24, 2017 at 19:05
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If $\alpha=0$, then $f_0(0)=1$ and $f_0(x)=0$ for $x \ne 0$.

Let $ \alpha >0$. Then:

$f_{\alpha}(x) = \dfrac{\sqrt{x}}{1-x}$ for $x \in [0, \alpha]$ and $f_{\alpha}(x) =0$ for $x \in \mathbb R \setminus [0, \alpha]$

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