Question

Find the value of $\displaystyle \sf \int_{0}^{ \infty } e {}^{ - {a}^{2} { {x}^{2} }^{} } dx$.
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Answer 1

$\textbf{Refer to the attachment for the solution!}$ $\textbf{Thank you!}$

Answer 2

Answer :

$\frac{ \sqrt{\pi} }{4a}$

Concept :

This is an improper integral. An improper integral is basically the area of the integrand calculated over an infinite range that may be [0, ∞), (-∞ , +∞) or (a , ∞) where a is any real number. Frankly speaking, it involves infinite limits of Integration.

So, in order to calculate these arbitrary integrals, you need to cleverly choose another integral (keeping the integrand same) involving different limits of integration of which atleast one should point to infinity. Now, this new integral must be chosen in such a way that after performing certain basic mathematical operations, you get you original integral back.

To calculate this integral in your question, my call is to choose another integral that is, integration of e^(-a²e²) from -∞ to +∞.

Now, I chose this integral coz the original integral, we wanna calculate, is half this integral. You see, integration is basically the process of calculating the area enclosed by a function. Now, closely look at the two attached images, the area of our new integral is twice the area of the original integral. Thus, all we need to do is calculate the new integral under the given limits and divide it by 2 to get the magnitude of the original integral.

Calculation :

The integral of the form e^(-ax²) from -∞ to +∞ is (√π)/(da/dx). Thus, integral of e^(-a²x²) from -∞ to +∞ will be (√π)/(2a). Now, the original integral is half this new integral, therefore, the magnitude of the original integral is (√π)/(4a).

Hence, integration of e^(-a²x²) from 0 to +∞ is (√π)/4a.