Calculate $$ \int_{[0,1]}(\log(x))^{2}x^{3/2}dx $$ with the help of the function $f:[0,1]\times[1,2] \rightarrow \mathbb{R}$, $f(x,t)=x^{t}$ by using the theorem of switching the differentiation and integration.
I found this exercise, but I don't quite understand the hint. How do I use the theorem? Can someone help me, please?
Let $g$ be the function defined by: $$g(t)=\int_0^1 f(x,t) dx ,\quad\forall t\in[1,2]$$
Then, $$g'(t)=\left(\int_0^1 f(x,t) dx\right)'=\int_0^1 f'(x,t) dx=\int_0^1 \log(x)x^tdx $$ $$g''(t)=\int_0^1(\log(x))^2 x^tdx$$ Now it is clear that, $$\int_0^1(\log(x))^2 x^{\frac{3}{2}}dx=g''(\frac{3}{2}) $$ But, $$g(t)=\int_0^1 f(x,t) dx=\left[\frac{1}{t+1}x^{t+1}\right]_0^1=\frac{1}{t+1} $$ Finally we have,$$ g'(t)=-\frac{1}{(t+1)^2}$$ $$g''(t)=\frac{2}{(t+1)^3} $$ $$g''(\frac{3}{2})=\frac{16}{125}$$
$$\int_{[0,1]}(log(x))^{2}x^{3/2}dx=\frac{16}{125}$$
$$\int_{0}^{1}\log^2(x) x^{\alpha}\,dx = \frac{d^2}{d\alpha^2}\int_{0}^{1}x^{\alpha}\,dx = \frac{d^2}{d\alpha^2}\frac{1}{1+\alpha} = \frac{2}{(1+\alpha)^3}$$ hence the answer is given by $\frac{2}{\left(1+\frac{3}{2}\right)^3}=\color{red}{\frac{16}{125}}$. I leave to you to justify why this actually works.