# Thread: help with this :(

1. ## help with this :(

let $
f(x) = \left[ {\int\limits_0^x {e^{ - t^2 } dt} } \right]^2 \,\,\,\,\,\,g(x) = \left[ {\int\limits_0^1 {e^{ - x^2 (1 + t^2 )} } \cdot \frac{1}{{(1 + t^2 )}}} \right]^2
$

show that $
f'(x) + g'(x) = 0
$

and show that $
f(x) + g(x) = \frac{\pi }{4}

$

finally prove that $
\int\limits_0^\infty {e^{ - t^2 } dt} = \frac{{\sqrt \pi }}{2}
$

i cant seem able to show that f'(x)+g'(x)=0, i think its beacause i cant derivate g(x). Help plz

2. first $g(x)$ was not meant to be defined as written, it's actually, without the square.

let $h(x,t)=\frac{e^{-x^{2}\left( 1+t^{2} \right)}}{1+t^{2}},$ and $\frac{\partial h}{\partial x}=-2xe^{-x^{2}\left( 1+t^{2} \right)},$ both continuous, then $g'(x)=-2x\int_{0}^{1}{e^{-x^{2}\left( 1+t^{2} \right)}\,dt},$ and $f'(x)=2e^{-x^2}\int_0^x e^{-t^2}\,dt,$ now substitute $t=ux$ in this integral and that shows that $f'(x)+g'(x)=0,$ thus $h(x)=f(x)+g(x)$ is constant and $h(0)=\frac\pi4$ and then $h(x)=\frac\pi4,$ obviously $\lim_{x\to\infty}h(x)=\frac\pi4,$ then $\lim_{x\to\infty}h(x)=\left(\int_0^\infty e^{-t^2}\,dt\right)^2$ and the conclusion follows.

3. A very nice way to find standart integral to be $\frac{\sqrt{\pi}}{2}$.
Could you help to understand that t=ux the sum of derivatives =0.
Thanks.