Question

Let $f,g :R-> R$ be to functions defined by

$f(x)=(\int_0^\infty \mathrm{e}^{-t^2}\,\mathrm{d}t)^2$ and $g(x)= \int_0^1\frac{e^\frac{-x^2(1+t^2)}{1}}{1+t^2}dt$

(a).Show that for all $x\in R :$  $f(x)+g(x)=\frac{\pi}{4}$

(b). Deduce that     $\int_0^\infty \mathrm{e}^{-t^2}\,\mathrm{d}t=\frac{\sqrt{\pi}{}}{2}$

 

Answer 1

solution

(a).Show that for all $x\in R :$ $f(x)+g(x)=\frac{\pi}{4}$

$=>f'(x)+g'(x)=2 (\int_0^x \mathrm{e}^{-t^2}\,\mathrm{d}t )' \int_0^x \mathrm{e}^{-t^2}\,\mathrm{d}t+ \int_0^1-2x\mathrm{e}^{-x^2(1+t^2)}\,\mathrm{d}t$

We have $(\int_0^x \mathrm{e}^{-t^2}\,\mathrm{d}t)'= \int_0^x 0dt + {e}^{-x^2}-e^2(0)={e}^{-x^2}$

$=> f'(x)+g'(x)=2{e}^{-x^2}\int_0^x \mathrm{e}^{-t^2}dt-2x{e}^{-x^2}\int_0^1 \mathrm{e}^{-x^2t^2}dt$  (1)

For $\int_0^1 \mathrm{e}^{-(xt)^2}dt$  , Let $u=xt=>du=xdt$ , Then $t=0=>u->0 , t=1=>u->x$

$=>\int_0^1 \mathrm{e}^{-(xt)^2}dt=\int_0^x \mathrm{e}^{-u^2}.\frac{1}{x}du=\frac{1}{x}\int_0^x \mathrm{e}^{-u^2}du$  (2)

(1)&(2) : $f'(x)+g'(x)=2{e}^{-x^2}\int_0^x \mathrm{e}^{-t^2}dt-2{e}^{-x^2}\int_0^x \mathrm{e}^{-u^2}du$

$=>f'(x)+g'(x)=0 =>f(x)+g(x)=K\in R$   (3)

we have $f(0)+g(0)=0 +\int_0^1 \frac{1}{1+t^2}dt=\left.arctan(t)\right|_0^1=\frac{\pi}{4}$  (4)

Therefore, $f(x)+g(x)=\frac{\pi}{4}$

 (b). Deduce that $\int_0^\infty \mathrm{e}^{-t^2}\,\mathrm{d}t=\frac{\sqrt{\pi}{}}{2}$

we have $f(x)+g(x)=\frac{\pi}{4}$ $<=>\lim\limits_{x \to +\infty} (f(x)+g(x))=\frac{\pi}{4}$

$<=>( \int_0^\infty \mathrm{e}^{-t^2}\,\mathrm{d}t)^2+0=\frac{\pi}{4}$

Therefore, $\int_0^\infty \mathrm{e}^{-t^2}\,\mathrm{d}t=\frac{\sqrt{\pi}{}}{2}$

(a).Show that for all $x\in R :$ $f(x)+g(x)=\frac{\pi}{4}$

Therefore, $f(x)+g(x)=\frac{\pi}{4}$

 (b). Deduce that $\int_0^\infty \mathrm{e}^{-t^2}\,\mathrm{d}t=\frac{\sqrt{\pi}{}}{2}$