The length of parametric curve defined by integrals A curve is defined by the parametric equations $$x=\int \limits_{1}^{t}\dfrac{\cos u}{u}du, \ y=\int \limits_{1}^{t}\dfrac{\sin u}{u}du.$$ Find the length of the arc of the curve from the origin to the nearest point where there is a vertical tangent line.
My solution: In order to find point where there is a vertical tangent line we need to solve $\dfrac{dx}{dt}=0$ and the nearest point is $\dfrac{\pi}{2}$. Using FTC we have $\sqrt{(x')^2+(y')^2}=\dfrac{1}{t}$. Hence, $$L=\int\limits_{0}^{\pi/2}\dfrac{dt}{t}=\ln t |_{0}^{\pi/2}$$
But $\ln(0)=-\infty$. Can anyone explain why there is a problem here?
 A: You have found the value where there is the nearest vertical tangent line, that is $t=\pi/2$. Also, you need the value of $t$ for which the curve passes through the origin: this is clearly $t=1$. Therefore, these are the limits in which you have to compute the arc lenght.
Following your computations and using these limits for $t$ we get
$$\ell=\int\limits_{1}^{\pi/2}\dfrac{dt}{t}=\ln t |_{1}^{\pi/2}=\ln \dfrac{\pi}{2}.$$
A: Worked it out this way:
$$ \text{slope}= \tan \phi=\frac{dy}{dx}=\frac{\sin u /u}{\cos u /u} = \tan u \tag 1$$
$$ \phi=u \tag 2 $$
$$ds^2= dx^2+dy^2 = \big(\frac{du}{u}\big)^2 \tag 3 $$
$$ ds= \frac{du}{u}, \text{integrate with BC.. } \;$$
Arc Length required
$$ s= log \frac {u}{u_i};\text {and further on..} u=u_i e^s \tag 4 $$
$$ \phi = \phi_i e^s \tag 5 $$
Differentiate for curvature with respect to arc
$$ \kappa = \phi_i e^s \tag 6$$
$$ \kappa = \phi \tag 7 $$
All the three are intrinsic/ natural differential equations.
Similar to Cornu spiral (clothoid) but is not the same. Maybe an exponential Cornu Spiral..
Numerical solution
phii = Pi/2; NDSolve[{PH'[s] == phi E^s, X'[s] == Cos[PH[s]], 
  Y'[s] == Sin[PH[s]], PH[0] == phii, X[0] == 0, Y[0] == 0}, {PH, X,  Y}, {s, 0, 3}];
{x[u_], y[u_], ph[u_]} = {X[u], Y[u], PH[u]} /. First[%];
ParametricPlot[{x[s], y[s]}, {s, 0, 3}, PlotStyle -> Thick, 
 GridLines -> Automatic]


Can be made to pass through origin by choice of BC.
