Question: Show that $\int^\infty_0 3^{-4x^2}dx = \frac{\sqrt{\pi }}{4\sqrt{log 3}}$
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Subject : Applied Mathematics 2

Topic : Numerical solution of ordinary differential equations of first order and first degree, Beta and Gamma Function

Difficulty: High

mumbai university • 68 views
 modified 9 weeks ago by Yashbeer ★ 160 written 9 weeks ago by Ankit Pandey ★ 60
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Solutions:

Let $\quad 1=\int_{0}^{\infty} 3^{-4 x^{2}} d x$

put $\quad 3^{-4 x^{2}}=e^{-t}$

taking log on both sides,

$4 x^{2} \log 3=t$

$x^{2}=\frac{t}{4 \log 3} \quad=\gt\quad x=\frac{\sqrt{t}}{2 \sqrt{\log 3}}$

diff w.r.t x,

$d x=\frac{t^{-1 / 2}}{4 \sqrt{\log 3}} d t$

$\therefore \mathrm{I}=\int_{0}^{\infty} \frac{e^{-t}}{4 \sqrt{\log 3}} t^{-1 / 2}$

$\therefore \mathrm{I}=\frac{1}{4 \sqrt{\log 3}} \int_{0}^{\infty} e^{-t} \cdot t^{-1 / 2} \mathrm{dt}$

$\therefore \mathrm{I}=\frac{\sqrt{\pi}}{4 \sqrt{\log 3}}$ $\ldots \ldots \ldots\left\{\int_{0}^{\infty} e^{-t} \cdot t^{-1 / 2} d t=\sqrt{\pi}\right\}$