Question

Let $I_1={\int }_0^x{e}^{tx}{e}^{-t^2}dt$ and $I_2={\int }_0^x{e}^{-t^2/4}dt$ 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)

$e^{x^2/4}$

$I_1={\int }_0^x{e}^{t(X-t)}dtt(x-t)=-[t^2-tx]=-[{(t-\frac{x}{2})}^2-\frac{x^2}{4}]=\frac{x^2}{4}-{(t-\frac{x}{2})}^2$

$I_1={\int }_0^x{e}^{\frac{x^2}{4}-{(t-\frac{x}{2})}^2}dt=e^{\frac{x^2}{4}}{\int }_0^x{e}^{-\frac{{(2t-x)}^2}{4}}dt;2t-x=y\Rightarrow dt=\frac{dy}{2}=\frac{e^{\frac{x^2}{4}}}{2}{\int }_{-x}^x{e}^{-}\frac{y^2}{4}dy$

$I_1=e^{\frac{x^2}{4}}{\int }_0^x{e}^{-}\frac{t^2}{4}dt=e^{\frac{x^2}{4}}.I_2\Rightarrow \frac{I_1}{I_2}=e^{\frac{x^2}{4}}$