I was trying to solve the above using log of integral. I maybe on an endless road of solving this question, but I wanted to know if such solution is possible. Here is my approach:
$$ I = \int e^{-x^2}dx $$ $$ I' = e^{-x^2} $$ Now taking $\ln$ on both the sides, $$\ln I' = -x^2$$ Now, differentiating both sides, $$ \frac{I''}{I'} = -2x $$ $$ I'' + 2xI' = 0 $$ With the initial conditions $I'(0) = 1$ and $I'(\infty) = 0$. I tried solving the above equation, but didn't succeed. Here are my approaches
- Case 1: Seeking solution in the form of $I(x) = x^\lambda$ $$ I' = \lambda x^{\lambda -1} $$ $$ I'' = \lambda (\lambda -1 )x^{\lambda - 2} $$ Putting the above in the differential equation $$ \lambda(\lambda - 1)x^{\lambda - 2} + 2x \lambda x^{\lambda - 1} = 0 $$ $$ x^\lambda[ \lambda (\lambda - 1)x^2 + 2\lambda] = 0 $$ Got $x^2$ term, so can't proceed further.
- Case 2 Seeking solution in the form $I = e^{\lambda x}$ $$ e^{\lambda x}[\lambda^2 + 2x \lambda] = 0 $$ Got $x$ term, so can't proceed further.
- Case 3 Using parameter $b$ here, $$ \int_0^bI'' dx = - \int_0^b 2xI' dx $$ $$I'(b) - I'(0) = -2{[xI(x)]_0^b + 2 \int_0^b I(x) dx}$$ Struck again.