Can $\int_{x=0}^{\infty} x\mathrm{e}^{-\alpha x^2}\,\mathrm {d}x$ be evaluated by parts to show that $\int_{x=0}^{\infty} x\mathrm{e}^{-\alpha x^2}\,\mathrm {d}x= \frac{1}{2\alpha}$

I know that this can be done without parts by means of the substitution $-\alpha x^2=u \Rightarrow -2\alpha x\, \mathrm {d}x=\mathrm {d}u$, then $x\mathrm dx=\frac {-1}{2\alpha}\mathrm du$; such that

$\displaystyle\int_{x=0}^{\infty} xe^{-\alpha x^2}\,dx=\frac {-1}{2\alpha}\displaystyle\int_{x=0}^{\infty} e^u \mathrm du= \left [ \frac {-1}{2\alpha}\displaystyle e^{-\alpha x^2} \right]_{x=0}^{\infty}=\frac{1}{2\alpha}$

However; Can it be done like this?

$\color{blue}{I}$ $=\displaystyle\int_{x=0}^{\infty} xe^{-\alpha x^2}\,dx= x\int_{x=0}^{\infty}e^{-\alpha x^2} - \int_{x=0}^{\infty}\left(\int_{x=0}^{\infty}e^{-\alpha x^2}\mathrm dx\right)\mathrm{d}x$

Now $$\int_{x=0}^{\infty}e^{-\alpha x^2}\mathrm dx= \frac{1}{2}\left(\frac{\pi}{\alpha}\right)^\frac{1}{2}\tag{1}$$

Now by insertion of $(1)$ into $\color{blue}{I}$ yields,

$\color{blue}{I}=\displaystyle\int_{x=0}^{\infty} xe^{-\alpha x^2}\,\mathrm{d}x=\underbrace{ \left [ x\frac{1}{2}\left(\frac{\pi}{\alpha}\right)^\frac{1}{2} \right ]_{x=0}^{\infty} - \int_{x=0}^{\infty} \frac{1}{2}\left(\frac{\pi}{\alpha}\right)^\frac{1}{2} \mathrm {d}x}_{\large\text{$\color{red}{\mathrm{Undefined}}$}}\ne \frac{1}{2\alpha}$ which is clearly a contradiction.

I realize that this may be blatantly obvious to many of you why this will never work, but it is not clear to me.

What I can't understand is that if asked to evaluate say $\int x\sin x\,\mathrm{d}x$ by parts I would do the following: $\int x\sin x\,\mathrm dx= x\int \sin x\,\mathrm dx-\int\left(\int \sin x\,\mathrm dx\right)\mathrm{d}x= \sin x -x\cos x$ + C. But the method does not work for $\int_{x=0}^{\infty} x\mathrm{e}^{-\alpha x^2}\,\mathrm dx$ by parts as shown above. Why is this?

Could someone please explain to me why I cannot perform this integral by parts to get the answer $\dfrac{1}{2\alpha}$?