I have a question about the following integration $\int_0^T \frac{\partial f(x+t)}{\partial x} dt$
Is it equal to F(x+T)-F(x)
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$\begingroup$ Why use partial derivative symbols? Did you mean $f(x,t)$? $\endgroup$– OFRBGMar 6, 2017 at 22:23
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$\begingroup$ no it is f(x+t) $\endgroup$– Isaac LiuMar 6, 2017 at 22:24
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1 Answer
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It equals $f(x+T) - f(x)$. $$ \frac{\partial}{\partial x}\left[f(x+t)\right] = f'(x+t) = \frac{\partial}{\partial t}\left[f(x+t)\right] $$ then use the fundamental theorem of calculus on the integral.
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$\begingroup$ I'm not the asker. How can you take two different partial derivatives and have the same answer? $\endgroup$– OFRBGMar 7, 2017 at 0:37
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$\begingroup$ @O.VonSeckendorff If it helps, consider the function $g(x,t) = f(x + t)$. Then for this particular function, it happens to be true that $\partial g/\partial x = \partial g/\partial t$. $\endgroup$ Mar 8, 2017 at 18:18