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I would like to differentiate $$ f(x) = \int_x^0 {\cos(xt) \over t} dt $$ with respect to x.

I tried to use the fundamental theorem of calculus, but the $xt$ inside the cosine is preventing me from proceeding. I suspected u-substitution by letting $u = xt$ and I had

$$\int_0^{x^2} {\cos(u) \over u} du$$

and then worked with integration by parts saying $\omega = \cos(u), dv = {du \over u}$, etc.

However, first of all I couldn't get the right solution, and this method took too much time when I have to solve this in less than 2-3 min because it is a GRE question.

Can someone help me out ?

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Your integral blows up at $0$. You cannot perform that integration at all. –  Pedro Tamaroff Apr 17 '13 at 21:04
    
Thanks for helping me out. It really means a lot. –  hyg17 Apr 17 '13 at 21:51
    
This type of problem -- differentiation of an integral function with a function as limit of integration -- often comes up on first-semester calculus final exams. Here's another recent question: math.stackexchange.com/questions/359729/… –  RecklessReckoner Apr 18 '13 at 2:56

2 Answers 2

up vote 1 down vote accepted

The integral does not exists, but let's suppose it is ok, the method will be the same.

Ok, the first thing to note is that when you change variables, you get in fact: $$ \int_{x^2}^0 \frac{cos(u)}{u} du $$ Now, if $F$ is the primitive of $cos(u)/u$ this integral gives: $$ F(0)-F(x^2) $$ If you differentiate it with chain's rule, you get: $$ -F'(x^2)\cdot2x $$ Because $F$ is the primitive that gives: $$ -2\frac{cos(x^2)}{x} $$

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If this is not yet the good answer, try to see if you put limits too early in the question. This would be a good reason for the integral to blow up, discarding an error from the part of the person that made the question. –  juanrapha Apr 17 '13 at 21:14
    
Great ! this was the answer that I was looking for. I knew that something was wrong... –  hyg17 Apr 17 '13 at 21:49

This integral does not converge. This can be seen from the Taylor expansion of $\frac{cos(xt)}{t}$, which has the first term of $\frac{1}{t}$. Integrating term by term (just looking at the first term) yields $\ln(t)\bigg|_{x}^{0}=\ln(x)-\ln(0)$, which is undefined. This $\ln(0)$ term is not canceled by another part of the Taylor series, so the integral must diverge.

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This is good, but it would me more appropriate to talk about $\lim\limits_{x\to 0^+}\log x$ and not "$\log 0$". –  Pedro Tamaroff Apr 17 '13 at 21:11
    
Yes, it is correct to use the limit as opposed to "ln(0)" in an improper integral. –  metsgiantsfan234 Apr 17 '13 at 21:14
    
Not just in an improper intgral, but everywhere. $\log x$ is defined only for $x>0$. –  Pedro Tamaroff Apr 17 '13 at 21:15
    
Thanks for the help ! –  hyg17 Apr 17 '13 at 21:51

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