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From below mentioned accepted answer, I am facing some further difficulties.
However, in $0\leq x \leq 1$ the area under PDF is coming as $0.9749$ and in the range of $1\leq x \leq 2$, when I am integrating
$\int_1^22 \arcsin \left( \frac 2 x - 1 \right)dx$ by substitution method $t=(\frac{2}{x}−1)$ and on proceeding we get
${\pi}-\int_0^1\frac{4}{(1+t)(\sqrt{(1-t^2)}}dt$, on integrating the following integral again by substitution of $t=sin\theta$, it is coming as
$\pi-\int_{\frac{\pi}{2}}^0\frac{4}{1+sin\theta}d\theta$, which is becoming divergent.
So, please tell me how the area under this PDF becomes $1$.
Your help will be appreciated.
Suppose the pdfs of $\hat X$ and $\hat Y$ are zero outside $(0, 1)$. The convolution of the pdfs can be split into two integrals: $$f_{\hat X + \hat Y}(x) = (f_{\hat X} * f_{\hat Y})(x) = \\ \int_0^x f_{\hat X}(\tau) f_{\hat Y}(x - \tau) d\tau \,[0 < x < 1] + \\ \int_{x - 1}^1 f_{\hat X}(\tau) f_{\hat Y}(x - \tau) d\tau \,[1 < x < 2].$$ If $f_{\hat X}(x) = f_{\hat Y}(x) = (1/\sqrt x - 1)[0 < x < 1]$, $$f_{\hat X + \hat Y}(x) = (x - 4 \sqrt x + \pi) [0 < x < 1] + \\ \left( 2 \arcsin \left( \frac 2 x - 1 \right) + 4 \sqrt {x - 1} - x - 2 \right) [1 < x < 2], \\ f_{X + Y}(x) = \frac 1 {a^2} f_{\hat X + \hat Y} \left( \frac x {a^2} \right).$$