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 November 29th, 2015, 07:58 PM #1 Newbie   Joined: Nov 2015 From: Philippines Posts: 1 Thanks: 0 Application of First Fundamental Theorem of Calculus Find the second derivative of f($x$)=$\displaystyle \int_{-\infty}^{x^2/2} e^{-(x^2+1)t^2} dt$ My solution: f($x$)=$\displaystyle \int_{-\infty}^0 e^{-(x^2+1)t^2}dt + \int_0^{x^2/2} e^{-(x^2+1)t^2}dt$ I may be able to find a closed form for the first integral $\displaystyle \int_{-\infty}^0 e^{-(x^2+1)t^2}dt$ by transforming into polar coordinates. My concern is the second integral: h($x$) = $\displaystyle \int_0^{x^2/2} e^{-(x^2+1)t^2}dt$ How do I apply Chain rule to get h'($x$)? Last edited by skipjack; November 30th, 2015 at 12:59 AM. Reason: to use latex
 November 30th, 2015, 01:05 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,270 Thanks: 1958 I doubt that the final answer can be given in closed form. I suggest that you make a substitution so that the original integral can be written in terms of the error function, erf(x).
 November 30th, 2015, 03:00 PM #3 Global Moderator   Joined: May 2007 Posts: 6,683 Thanks: 658 There is no point in splitting the integral. If you apply the basic definition of the derivative, $\displaystyle f'(x)=\lim_{h ->0}\frac{f(x+h)-f(x)}{h}$, you will be able to work it out by brute force. For your problem let $\displaystyle f(x)=\int_{-\infty}^{g(x)}p(x,t)dt$ $\displaystyle f'(x)=p(x,g(x))g'(x)+\int_{-\infty}^{g(x)}\frac{\partial p(x,t)}{\partial x}dt$ I'll let you work out the second derivative.

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