0
1.3kviews
Show that $\int_{0}^{\infty}7^{-4x^2}dx=\frac{\sqrt{\pi}}{4\sqrt{log\,7}}$ .
1 Answer
0
52views

Put,

$7^{-4x^2}= e^-t \hspace{0.3cm}=\gt\hspace{0.3cm}-4x^2log7=-t$

$x^2=\frac{t}{4log7} \hspace{0.3cm}=\gt\hspace{0.3cm} x=\frac {t^{1/2}}{2\sqrt{log7}}$

$\hspace{2cm}=\gt dx=\frac{1}{2}\frac{t^{-1/2}}{\sqrt{log7}}dt$

$\int_{0}^{\infty}e^{-t}\frac{1}{2}\frac{t^{-1/2}}{\sqrt{log7}}dx=\frac{1}{2\sqrt{log7}}\int_{0}^{\infty}e^{-t}t^{-1/2}dt$

$\hspace{3cm}=\frac{1}{2\sqrt{log7}}\sqrt{1/2} = \frac{\sqrt{\pi}}{2\sqrt{log7}}$

Please log in to add an answer.