for any $k\in N$, p.v$\int_{a}^{b}\frac{\cos kx}{t-x}dx=?$

Question

We know that the Hilbert transform of cosine function is sine,see http://mathworld.wolfram.com/HilbertTransform.html.

Now, we don't integral from $-\infty \to \infty$. We integral from $a \to b(0<a<b)$. My question is:

For any $k\in N$,

$$p.v.\int_{a}^{b}\frac{\cos kx}{t-x}dx=?$$

where the Cauchy principal value is taken in the integrals.

Answer 1

The Maple command $$int(cos(k*x)/(t-x), x = a .. b, CauchyPrincipalValue)\,assuming \,t > a, t <b, k >0 $$ outputs $$-{\it Si} \left( ak-tk \right) \sin \left( tk \right) +\cos \left( tk \right) {\it Ci} \left( -ak+tk \right) +{\it Si} \left( bk-tk \right) \sin \left( tk \right) -{\it Ci} \left( bk-tk \right) \cos \left( tk \right) .$$ See Si for info.

Answer 2

Assuming $a < t < b$, just do $$ \text{PV}\int_{-\infty}^\infty \frac{cos(kx)}{(x-t)} \, dx - \int_{-\infty}^a \frac{cos(kx)}{(x-t)} \, dx - \int_{b}^\infty \frac{cos(kx)}{(x-t)} \, dx .$$ And if you don't have $a\le t\le b$, then its not a principle value. (And if $t=a$ or $t=b$ then the integral is some kind of infinite value unless by chance $\cos(ka)$ or $\cos(kb)$ happen to be zero.)

And of course Si and Ci are just names for integrals we don't know how to evaluate (namely $\sin(x)/x$ and $\cos(x)/x$). So they are bound to appear in the answer.