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Find the derivative of $f(x)= \int_x^0 \frac{\cos(xt)}{t} dt$.

My first reaction was to apply the FTOC, but I don't believe I can do this because $\frac{\cos(xt)}{t}$ is not defined at $t=0$ and thus it is not continuous in the interval of integration. I am trying to re-learn calculus after being away from it for so long and I am a bit rusty. My book doesn't give a clear solution, so any pointers on what to do in this situation would be greatly appreciated.

Edit: After some internet digging, I've learned that what I need to solve the problem is differetiation under the integral sign, which states $$\frac{d}{dx}(\int_{a(x)}^{b(x)}f(x,t) dt)=f(x,b(x))\cdot b'(x) - f(x,a(x))\cdot a'(x)+\int_{a(x)}^{b(x)}\frac{\partial}{\partial x}f(x,t)dt $$

Thus, $\frac{d}{dx}(\int_x^0 \frac{\cos(xt)}{t} dt)=-\frac{\cos(x^2)}{x}+\int_0^x \sin(xt)dt=-\frac{\cos(x^2)}{x}-\frac{\cos(x^2)}{x}+\frac{1}{x}=\frac{1}{x}(1-2\cos(x^2))$

Edit 2: I'm not so sure that my above method is correct anymore.

Poko
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    $f(x)$ isn't well defined. The integral does not converge. (Sorry for the irritating typo in the first comment). – user251257 Aug 26 '15 at 00:39
  • Although the integral itself doesn't converge, we can still solve for the derivative of the integral (see my edit for details). – Poko Aug 26 '15 at 01:47
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    blindly applying some formula won't help you. $f(x,0)$ isn't defined nor exists as a limit – user251257 Aug 26 '15 at 01:47
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    notice that $\cos$ is roughly $1$ near zero. So you are basically trying to derive $\int_x^0 \frac{1}{t} ;\mathrm d t$. The integral is $-\infty$ for any $x\ne 0$. What should the derivative of $-\infty$ be? – user251257 Aug 26 '15 at 01:57
  • What you are saying indeed makes sense, but I am still skeptical. I see that $\frac{\cos(xt)}{t}$ is not defined at $t=0$, which was my original concern. However both my book and wolfram alpha will give me the same answer that I derived by blindly applying the formula. I'm clearly missing a technique of integration. – Poko Aug 26 '15 at 04:16
  • user251257 is right. For more to be said, you'd need to quote the book and/or specify the Wolfram|Alpha input that appear to be giving the incorrect result. – joriki Aug 26 '15 at 06:12
  • Wolfram|Alpha input. – Poko Aug 26 '15 at 06:36
  • well, WA thinks the derivative of $\int_x^0 1/t ; dt$ is $-1/x$... – user251257 Aug 26 '15 at 11:55
  • Ok, thanks for your time. – Poko Aug 26 '15 at 18:00
  • Note that underneath the incorrect result, Wolfram|Alpha also says (contradicting itself) "integral does not converge". – joriki Aug 27 '15 at 14:08

1 Answers1

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As mentioned in the comments, this integral does not converge for any $x \neq 0$. This is due to the fact that $$ \int_0^{a} \frac{1}{x} dx$$ does not converge for any $a$. So it does not make sense to talk about the derivative of the integral in your question.

We can note that if the original integral was $$ \int_x^{1} \frac{\cos tx}{t} dt,$$ then this has a derivative that makes sense for all $x > 0$. Further, the derivative is exactly what you think the derivative should be, as indicated by the work in the OP.

davidlowryduda
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