How this integral is calculated

encada

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Hi everyone,
I tried every method I know but I can't find the same result for this integral from a paper. the closest I got is the same with a factor of 0.5 instead of 0.44, when using binomial expansion approximation of the squared term. the f(alpha) is independent from x. I would be very grateful if somebody enlighten me....
 

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The "binomial expansion" is \(\displaystyle (x+ y)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i \end{pmatrix} x^iy^{n-i}\) where n is a positive integer and \(\displaystyle \begin{pmatrix} n \\ i \end{pmatrix}= \frac{n!}{i!(n-i)!}\) is the "binomial coefficient".

The "generalized binomial expansion" where n does NOT have to be a positive integer, replaces the binomial coefficient with \(\displaystyle \frac{r(r- 1)(r- 2)\cdot\cdot\cdot (r-i)}{i}\) and the finite sum becomes an infinite series. In particular \(\displaystyle (1+ x)^{1/2}= 1+ \frac{1}{2}x- \frac{1}{8}x^2+ \frac{1}{16}x^3\cdot\cdot\cdot\).
 
The "binomial expansion" is \(\displaystyle (x+ y)^n= \sum_{i=0}^n \begin{pmatrix}n \\ i \end{pmatrix} x^iy^{n-i}\) where n is a positive integer and \(\displaystyle \begin{pmatrix} n \\ i \end{pmatrix}= \frac{n!}{i!(n-i)!}\) is the "binomial coefficient".

The "generalized binomial expansion" where n does NOT have to be a positive integer, replaces the binomial coefficient with \(\displaystyle \frac{r(r- 1)(r- 2)\cdot\cdot\cdot (r-i)}{i}\) and the finite sum becomes an infinite series. In particular \(\displaystyle (1+ x)^{1/2}= 1+ \frac{1}{2}x- \frac{1}{8}x^2+ \frac{1}{16}x^3\cdot\cdot\cdot\).
thank you very much for answering. this is what I have done using the expansion up to the second term. taking more terms doesn't look to bring the result closer to that in the paper. I was trying to figure out how to type my solution directly on the post but didn't know how, so i posted a picture
 

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