# How do you compute the derivative of $\int_x^0\frac{\cos(xt)}{t}dt$?

What is the derivative of $\int_x^0\frac{\cos(xt)}{t}dt$ with respect to $x$? Using Leibniz' rule, I think it equals \begin{align} -\frac{\cos(x^2)}{x}+\int_x^0 -\sin(xt)dt &= -\frac{\cos(x^2)}{x}+\frac{\cos(xt)}{x}\bigg\vert_x^0 \\ &= \frac{1}{x}(1-2\cos(x^2)) \end{align}

Is that all there is to it? I'm doubtful since Wolframalpha says something about the integral not converging.

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Before even talking about the derivative of a function, you need to check wether the function makes sense.

Let $$f(t)=\frac{\cos(tx)}{t}$$

Then $f$ is defined for every $t$ except $t=0$. If we want to start talking about

$$\int\limits_0^x {f(t)}dt = \int\limits_0^x {\frac{{\cos (tx)}}{t}}dt$$

we thus have to take extra care about what happens near $0$. Clearly, we want to consider an improper integral. The function $f(t)$ is continuous on any $[\epsilon,x]$ for $x,\epsilon>0$, thus integrable on $[\epsilon,x]$, so we want to look at $\epsilon \to 0^+$.

$$\int\limits_\epsilon ^x {f(t)}\,dt = \int\limits_\epsilon ^x {\frac{{\cos (tx)}}{t}} \, dt$$

But

$$\frac{{\cos (tx)}}{t} = \frac{1}{t} - x^2\frac{{t{}}}{{2}} + o\left( {{t}} \right)$$

so,

$$\int\limits_\epsilon ^x {\frac{{\cos (tx)}}{t}} = \log x \color{red}{- \log \epsilon } - {x^2}\frac{{{x^2}}}{4} + {x^2}\frac{{{\epsilon ^2}}}{4} + o(t^2)$$

Do you see now what's the problem when $\epsilon\to 0^+$?

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Ah, so the whole thing will blow up towards $+\infty$. –  Noomi Holloway Sep 30 '12 at 2:37
Yes. More succinctly, $$\frac{{\cos (tx)}}{t} \sim \frac{1}{t}$$ for $t\to 0$. –  Pedro Tamaroff Sep 30 '12 at 2:45
I don't think it converges. The integrand goes like $t^{-1}$ near $t=0$, so would integrate to a lograrithmic divergence.