Answer :

Solution :

${\int }_{-\infty }^{\infty }{e}^{-a(x+b{)}^2}dx$

substitute, x + b = t

dx = dt

at x = -∞ , t = -∞

x = ∞ , t = ∞

${\int }_{-\infty }^{\infty }{e}^{-a(x+b{)}^2}dx$ = ${\int }_{-\infty }^{\infty }{e}^{-a{t}^2}dt$    ... (1)

Let substitute, at² = y² ⇒ $t=\frac{y}{\sqrt{a}}$

on differentiating at² = y²

2 at dt = 2ydy

dt = $\frac{ydy}{a\frac{y}{\sqrt{a}}}$ = $\frac{ydy}{a\frac{y}{\sqrt{a}}}$ = $\frac{dy}{\sqrt{a}}$

Now substitute dt in (1)

${\int }_{-\infty }^{\infty }{e}^{-a{t}^2}dt$ = $\frac{1}{\sqrt{a}}{\int }_{-\infty }^{\infty }{e}^{-y^2}dy$

It is given that ${\int }_{-\infty }^{\infty }{e}^{-x^2}dx=\sqrt{\pi }$

Hence we get

$\frac{1}{\sqrt{a}}{\int }_{-\infty }^{\infty }{e}^{-y^2}dy$ = $\sqrt{\frac{\pi }{a}}$