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Show that $\int_{0}^{\infty}7^{-4x^2}dx=\frac{\sqrt{\pi}}{4\sqrt{log\,7}}$ .
1 Answer
| written 8.0 years ago by | • modified 8.0 years ago |
Put,
$7^{-4x^2}= e^-t \hspace{0.3cm}=\gt\hspace{0.3cm}-4x^2log7=-t$
$x^2=\frac{t}{4log7} \hspace{0.3cm}=\gt\hspace{0.3cm} x=\frac {t^{1/2}}{2\sqrt{log7}}$
$\hspace{2cm}=\gt dx=\frac{1}{2}\frac{t^{-1/2}}{\sqrt{log7}}dt$
$\int_{0}^{\infty}e^{-t}\frac{1}{2}\frac{t^{-1/2}}{\sqrt{log7}}dx=\frac{1}{2\sqrt{log7}}\int_{0}^{\infty}e^{-t}t^{-1/2}dt$
$\hspace{3cm}=\frac{1}{2\sqrt{log7}}\sqrt{1/2} = \frac{\sqrt{\pi}}{2\sqrt{log7}}$