The following substitution came up in the solution set when I was practicing Diff. Geom., and it seems wrong to me: $$\int_0^t\cos(\ln x) dx = \int_0^{\ln(t)} \cos(u)e^u du$$ When $u$ is evaluated at the lower bound, it is not defined! Additionally, the limit $\ln s\downarrow 0 = -\infty$.
Is this just very sloppy notation in the solution set, or am I missing something else?
You are correct: $$x\in[0,t],\,u=\ln(x)\Rightarrow u\in(-\infty,\ln t]$$ and so: $$\int_0^t\cos(\ln(x))dx=\int_{-\infty}^{\ln(t)}e^u\cos(u)du$$ you may want to split this integral up into two parts: $I=I_1+I_2$ if that is clearer: \begin{align}I=\int_{-\infty}^0e^u\cos(u)du=\Re\int_{-\infty}^{\ln(t)}e^{(1+i)u}du=\Re\left[\frac{e^{(1+i)u}}{1+i}\right]_{-\infty}^{\ln(t)}\end{align} \begin{align}=\Re\left(\frac{(1-i)e^{(1+i)\ln(t)}}{2}\right)=\Re\left(\frac{(1-i)t\left(\cos(\ln(t))+i\sin(\ln(t))\right)}{2}\right)\end{align} $$=\frac t2\Re\left(\left[\cos(\ln(t))+\sin(\ln(t))\right]+i\left[\sin(\ln(t))-\cos(\ln(t))\right]\right)$$ $$=\frac t2\left[\cos(\ln(t))+\sin(\ln(t))\right]=\frac t{\sqrt{2}}\cos\left(\ln(t)-\frac{\pi}{4}\right)$$