GATE 2022 ME (ME2) – Question 14
Given $\int_{-\infty}^{\infty}e^{-x^2}\,dx=\sqrt\pi$. If $a$ and $b$ are positive integers, the value of $\int_{-\infty}^{\infty}e^{-a(x+b)^2}\,dx$ is ________.
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (B) $\sqrt{\pi/a}$
Explanation
**Step 1: substitute.** Let $t=\sqrt a\,(x+b)$. Then $x+b=t/\sqrt a$ and $dx=dt/\sqrt a$.
**Step 2: check the limits.** As $x\to\pm\infty$, $t\to\pm\infty$ (because $a>0$), so the limits do not change.
**Step 3: rewrite the integral.** Since $a(x+b)^2=t^2$:
$$\int_{-\infty}^{\infty}e^{-a(x+b)^2}dx=\frac1{\sqrt a}\int_{-\infty}^{\infty}e^{-t^2}dt=\frac{\sqrt\pi}{\sqrt a}=\sqrt{\frac{\pi}{a}}.$$
The shift $b$ drops out because the integral runs over the whole line. Answer **B**.