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November 29th, 2015, 07:58 PM   #1
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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
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November 30th, 2015, 01:05 AM   #2
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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).
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November 30th, 2015, 03:00 PM   #3
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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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