Question

Let $I_1=\underset{0}{\overset{x}{\int }}{e}^{tx}\cdot e^{-t^2}dt$ and $I_2=\underset{0}{\overset{x}{\int }}{e}^{-t^2/4}dt$ where $x>0$ then the value of $\frac{I_1}{I_2}$ is

• A. $e^{-x^2/2}$
• B. $e^{x^2/4}$
• C. $e^{-x^2/4}$
• D. $e^{x^2/2}$

Answer 1

Answer: (B)

$I_1=\underset{0}{\overset{x}{\int }}{e}^{t(x-t)}dt$ $t(x-t)=-\left[t^2-tx\right]=-\left[{\left(t-\frac{x}{2}\right)}^2-\frac{x^2}{4}\right]=\frac{x^2}{4}-{\left(t-\frac{x}{2}\right)}^2$ $I_1=\underset{0}{\overset{x}{\int }}{e}^{\frac{x^2}{4}{\left(t-\frac{x}{2}\right)}^2}dt=e^{\frac{x^2}{4}}\underset{0}{\overset{x}{\int }}{e}^{-\frac{(2t-x{)}^2}{4}}dt;2t-x=y\Rightarrow dt=\frac{dy}{2}$ $=\frac{e^{x^2/4}}{2}\underset{-x}{\overset{x}{\int }}{e}^{-\frac{y^2}{4}}dy$ $I_1=e^{\frac{x^2}{4}}\underset{0}{\overset{x}{\int }}{e}^{-\frac{t^2}{4}}dt=e^{\frac{x^2}{4}}\cdot I_2\Rightarrow \frac{I_1}{I_2}=e^{\frac{x^2}{4}}$