We have $\left \{ X(t),Y(t) \right \}, 1\leq t\leq \pi$ defined by the following two definite integral:
$X(t)=\int_{1}^{t}\frac{\cos z}{z^2}dz, Y(t)=\int_{1}^{t}\frac{\sin z}{z^2}dz$
$L$ be the length of the arc of the curve from the origin to the point $P$ on the curve at which the tangent is perpendicular to the x-axis. Then, the $L$ will be equal to?
I took the derivative of both $X(t),Y(t)$ using the Leibniz rule and then found the $\frac{dy}{dx}$ as follows:
$\frac{dy}{dx}=\tan (t)$
I am thinking of using the following formula for the length of the curve: $\int_{0}^{a}\sqrt{1+f'(x)^2} dx$
But I am unable to find what will be the upper limit of the above integral. Any help to find the upper limit will be appreciated. Thanks in advance
I find that doing this analysis in the complex plane simplifies it somewhat. First let me alter the notation in order to avoid confusion with the complex varaible $z$. To that end, let's write
$$ X(t)=\int_{1}^{t}\frac{\cos u}{u^2}du\\ Y(t)=\int_{1}^{t}\frac{\sin u}{u^2}du $$
Now we can express the complex variable as
$$z(t)=\int_{1}^{t}\frac{e^{iu}}{u^2}du$$
The arc length is given as follows in the complex plane,
$$s=\int |\dot z| dt$$
From Leibniz's rule
$$ \dot z=\frac{e^{it}}{t^2}\\ |\dot z|=\frac{1}{t^2} $$
And finally,
$$s=\int_1^T \frac{1}{t^2} dt=1-\frac{1}{T}$$
I have verified this result numerically (in comparison with the integral). Also note that WolframAlpha gives $\int_1^{\infty} 1/t^2 dt=1$.