Question

How would I take the derivative of the integral from 0 to x of Cos(x t)/t dt. How can I use the second fundamental theorem of calculus if there is an x inside the cos function?

Answer 1

is the derivative with respect to t

Answer 2

No, the derivative is wrt to x

Answer 3

\[ \frac{ d }{ dx}\int\limits_{0}^{x}\frac{Cos(x t)}{t}dt\]

Answer 4

the integral does not converge

Answer 5

First off, this integral Does converge, and there is nothing meaningless about the problem. I am taking the integral of a function of x, so x can appear inside and in the limits of the integral without a problem. I know the solution to the problem is
\[\frac{ 1}{ x }[2Cos(x^2)-1]\]
(you can easily check this on Wolfram Alpha) I'm just not exactly sure how to come this solution. Any help you can offer will be very appreciated.

Answer 6

use fundamental theorem of calculus.
http://www.sosmath.com/calculus/integ/integ03/integ03.html let f(t) = cos(tx)/t
\[\frac{ d }{ dx}\int\limits_{0}^{x}\frac{\cos(x t)}{t}dt = \cos(x^2)/x - \cos(0)/0\]

Answer 7

also W|A gives you result that computational result does not converge.
http://www.wolframalpha.com/input/?i=D%5BIntegrate%5BCos%5Bx+t%5D%2Ft%2C%7Bt%2C+0%2Cx%7D%5D%2C%7Bx%2C1%7D%5D

Answer 8

You are still not taking into account the factor of x inside the integral, simply using the fundamental theorem like you suggest does not provide the correct solution, since your calculation above does not reduce to the correct answer. Also, the fact that wolfram alpha does not "computationally converge" is meaningless considering we don't expect a numerical answer, we are looking for the answer to be a function of x. The first answer provided by wolfram alpha is what we want. I don't think the answer to this problem is as simple as you think it is.

Answer 9

There is an extension for multivariable functions.

Answer 10

http://en.wikipedia.org/wiki/Leibniz_integral_rule

Answer 11

\[
\frac{d}{dx}\int_a^{g(x)}f(x,t)\;dt = \int_a^{g(x)} \frac{\partial}{\partial x} f(x,t)\;dt+f(g(x),t)\frac{dg}{dx}
\]

Answer 12

@Dsmith3778 You can do it now.

Answer 13

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Answer 14

That is it! I was not aware of the extension for multivariable functions, thank you for your help @wio !