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