This is a GRE prep question: What's the derivative of $f(x)=\int_x^0 \frac{\cos xt}{t}\mathrm{d}t$?
The answer is $\frac{1}{x}[1-2\cos x^2]$. I guess this has something to do with the first fundamental theorem of calculus but I'm not sure how to use that to solve this problem.
Thanks in advance.
First Foundamental Theorem of Calculus : $$\frac{d}{dx}\int_{{\alpha(x)}}^{{\beta(x)}}f(x,y)\,dy=\int_{{\alpha(x)}}^{{% \beta(x)}}\frac{\partial f(x,y)}{\partial x}\,dy+\frac{d\beta(x)}{dx}f(x,\beta% (x))-\frac{d\alpha(x)}{dx}f(x,\alpha(x))\, $$ where : $$\alpha(x)=0, \beta(x)=x, f(x,y)=\frac{\cos xy}{y}$$
Note that not only the domain of integration but also the integrand depend on $x$. Let's (for a moment) write $F(x) = \int_x^0 f(x, t)\,dt = -\int_0^x f(x,t)\,dt$ for your integral. To handle the "double" $x$-dependence, we write $F$ as the composition of $\Delta\colon \mathbb R \to \mathbb R^2$, $x\mapsto (x,x)$ and $\Phi\colon \mathbb R^2 \to \mathbb R$, $(x_1, x_2) \mapsto -\int_0^{x_1} f(x_2, t)\, dt$. We have $F = \Phi \circ \Delta$, hence $$ F'(x) = \nabla \Phi \bigl(\Delta(x)\bigr) \cdot \Delta'(x) $$ by the chain rule. Now $\Delta'(x) = (1,1)$, and \begin{align*} \partial_1\Phi(x_1, x_2) &= -f(x_2, x_1)\\ \partial_2\Phi(x_1, x_2) &= -\int_0^{x_1} \partial_{x_2} f(x_2, t)\, dt \end{align*} (for the first derivative we used the fundamental theorem). Plugin everything together, we obtain $$ F'(x) = -f(x,x) - \int_0^x \partial_x f(x,t)\, dt $$
The integral does not exist, consequently it is not differentiable.
The integral does not exist, because for each $x$ ($x>0$, the case of negative $x$ is dealt with similarly) there is some $\varepsilon > 0$ such that $\cos(xt)> 1/2$ for all $t$ in the $t$-range $0\le t \le \varepsilon$. If one splits the integral $\int_x^0 = \int_\varepsilon^0 + \int_x^\varepsilon$, then the second integral exists, because the integrand is continuous in the $t$-range $\varepsilon \le t\le x $. The first is by definition $\lim_{\eta\rightarrow 0+}\int_\varepsilon^\eta$ and larger than $(-\ln \eta + \ln \varepsilon)/2$. So the limit does not exist.
