Given ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$

If a and b are positive integers, the value of ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-a(x+b{)}^2}dx$ is _________.

The correct option is:

$\sqrt{\frac{\pi }{a}}$

To find the value of the given integral, we can use the method of substitution to transform it into the standard Gaussian integral form.

Let the given integral be represented as $I$:

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

We define a new variable $u$ to simplify the exponent of the exponential function. Let us substitute:

$u=\sqrt{a}(x+b)$

Now, we find the differential $du$ by differentiating $u$ with respect to $x$:
$du=\sqrt{a}dx$
This allows us to write:
$dx=\frac{du}{\sqrt{a}}$

Next, we determine the new limits of integration. Since $a$ is a positive integer, $\sqrt{a}$ is also positive. Therefore:
As $x\to -\infty$, we have $u\to -\infty$.
As $x\to \infty$, we have $u\to \infty$.
Thus, the limits of integration remain unchanged.

Substituting these values into our integral, we get:

$I={\int }_{-\infty }^{\infty }{e}^{-u^2}\frac{du}{\sqrt{a}}$

Since $\sqrt{a}$ is a constant with respect to the variable of integration $u$, we can pull it outside the integral:

$I=\frac{1}{\sqrt{a}}{\int }_{-\infty }^{\infty }{e}^{-u^2}du$

We are given the standard Gaussian integral evaluation:

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

Using this identity for our variable $u$, we substitute ${\int }_{-\infty }^{\infty }{e}^{-u^2}du=\sqrt{\pi }$ back into the expression:

$I=\frac{1}{\sqrt{a}}\cdot \sqrt{\pi }=\sqrt{\frac{\pi }{a}}$