Question

Let a, b be positive real numbers such that \(a\lt b\). Given that
\(\lim\limits_{N\rightarrow\infin}\displaystyle\int^{N}_{0}e^{-t^2}dt=\frac{\sqrt \pi}{2},\) the value of
\(\lim\limits_{N\rightarrow\infin}\displaystyle\int^{N}_{0}\frac{1}{t^2}(e^{-at^2}-e^{-bt^2})dt\) is equal to

• A. √π(√a – √b).
• B. √π(√a + √b).
• C. -√π(√a + √b).
• D. √π(√b – √a).

Answer 1

The Correct Option is D

To solve this problem, we need to evaluate the limit:

\(\lim\limits_{N\rightarrow\infty}\displaystyle\int^{N}_{0}\frac{1}{t^2}(e^{-at^2}-e^{-bt^2})dt\)

Given:
- \(\lim\limits_{N\rightarrow\infty}\displaystyle\int^{N}_{0}e^{-t^2}dt=\frac{\sqrt \pi}{2}\)

First, let's consider the integral \(\int_{0}^{N} e^{-ct^2} \, dt\) for any positive constant \(c\) using the substitution \(u = \sqrt{c}t\), implying \(du = \sqrt{c}\, dt\). Therefore, \(dt = \frac{du}{\sqrt{c}}\) and \(t^2 = \frac{u^2}{c}\).

The integral becomes:

\(\int_0^{\sqrt{c}N} e^{-u^2} \frac{du}{\sqrt{c}} = \frac{1}{\sqrt{c}} \int_0^{\sqrt{c}N} e^{-u^2} du\)

As \(N \to \infty\), \(\sqrt{c}N \to \infty\) and we have:

\(\frac{1}{\sqrt{c}} \lim_{N \to \infty} \int_0^{\sqrt{c}N} e^{-u^2} du = \frac{1}{\sqrt{c}} \cdot \frac{\sqrt{\pi}}{2}\)

Considering the original limit:

\(\lim\limits_{N\rightarrow\infty}\displaystyle\int^{N}_{0}\frac{1}{t^2}(e^{-at^2}-e^{-bt^2})dt\)

We split it into two parts:

\(\lim\limits_{N\rightarrow\infty}( \int^{N}_{0}\frac{1}{t^2}e^{-at^2}dt - \int^{N}_{0}\frac{1}{t^2}e^{-bt^2}dt)\)

Evaluating both, using similar substitution as above, leads to:

\(\frac{1}{\sqrt{a}} \cdot \frac{\sqrt{\pi}}{2} - \frac{1}{\sqrt{b}} \cdot \frac{\sqrt{\pi}}{2} = \frac{\sqrt{\pi}}{2} \left( \frac{1}{\sqrt{a}} - \frac{1}{\sqrt{b}} \right)\)

Factoring the terms gives:

=\(\frac{\sqrt{\pi}}{2}\left( \frac{\sqrt{b} - \sqrt{a}}{\sqrt{ab}} \right) \cdot 2\sqrt{ab} = \sqrt{\pi} ( \sqrt{b} - \sqrt{a})\)

Therefore, the value of the limit is:

Option: \(\sqrt{\pi}(\sqrt{b}-\sqrt{a})\)