Suppose $$f(x)=\left(\int_{0}^{x}e^{-t^{2}}dt \right)^2 $$ and $$ g(x)=\int_{0}^{1}\frac{e^{-x^{2}{(t^{2}+1)}}}{t^{2}+1}dt. $$ 1) Find $f'(x)+g'(x)$.
2) Show that $f(x)+g(x)=\frac{\pi}{2}$.
I have no idea how to deal with this kind of questions. Any advice is appreciated.
The exam in which this question appeared is already over. I just want to learn how to deal with this type of question.
Let $h$ be the function $\Bbb R\to\Bbb R$, $$ h(x) = \sqrt{f(x)}=\int_{0}^{x}e^{-t^2}\; dt \ . $$ Then we have: $$ \begin{aligned} f'(x) &= 2h(x)\; h'(x)=2h(x)\cdot e^{-x^2}\ ,\\ g'(x) &= \frac{\partial}{\partial x} \int_{0}^{1}\frac{e^{-x^2(t^{2}+1)}}{t^2+1}\;dt \\ &= \int_0^1 \frac{\partial}{\partial x} \frac{e^{-x^2(t^{2}+1)}}{t^2+1}\;dt \\ &= \int_0^1 -2x\;e^{-x^2(t^2+1)}\;dt = -2\;e^{-x^2}\;\int_{t\in[0,1]} e^{-(tx)^2}\;\underbrace{x\; dt}_{d(tx)} \\ &\qquad\text{ Substitution: }u=tx\ ,\ du=x\; dt\ ,\ u\in[0,x]\ , \\ &=-2\; e^{-x^2}\int_{u\in[0,x]}e^{-u^2}\; du=-2e^{-x^2}\; h(x)\ . \\[3mm] &\qquad\text{ This implies: } \\ f'(x)+g'(x) &= 0\ . \end{aligned} $$ So the function $f+g$ is a constant function. We only need to compute it in a special point, say in zero. Of course $f(0)=h(0)^2=0^2=0$, and for $g(0)$ we have to compute $$ g(0) =\int_0^1 \frac{e^{-0^2(t^{2}+1)}}{t^2+1}\;dt =\int_0^1 \frac1{t^2+1}\;dt\ . $$ Which is the value of the last integral?!
