What is a solution to this integral? What is a solution to the following integral:
$$\int_0^\infty \, \frac{\cos{kt}}{\pi}\,\mathrm{d}k\,?$$
I have tried to evaluate this in the usual way:
$$\begin{align}\frac{1}{\pi}\int_0^\infty\,\cos{kt}\,\mathrm{d}k &= \frac{1}{\pi t} \left[\sin{k t}\right]\bigg|_{k\rightarrow 0}^{k\rightarrow \infty} \\ &=\frac{1}{\pi t} \left[\lim_{k\rightarrow \infty}\sin{k t} - \sin 0\right]  \\ &=\frac{1}{\pi t} \lim_{k\rightarrow \infty}\sin{k t},\end{align}$$
but $\lim_{k\rightarrow \infty}\sin{k t}$ doesn't converge within infinity. 
What is the trick here? Note that this is a homework question, so I don't need the complete solution, just directions where to start.
 A: The integral doesn't converge to any finite value. A graph of the function you're integrating would be a transformed version of $\cos k$, which oscillates back and forth. The area underneath would also oscillate back and forth, which is what your answer already tells you.
A: To answer your question, there is no finite value for the integral.
As you showed, the integral simplifies to the value:
$$\frac{1}{\pi t} \lim_{k\rightarrow \infty}\sin{k t}$$
Looking at the limit, as $k$ approaches infinity, the function $\sin{(kt)}$ will oscillate between the values $-1$ and $1$, and therefore won't converge to a finite value.
You could use taylor series of $\sin x$ to get a value through approximation, but the integral itself doesn't converge to any value.
A: I have found that a solution can can be inferred from equation 2.3.36 (pg. 137) in Dubin's Numerical and Analytical Methods for Scientists and Engineers using Mathematica, which finds that
$$\lim_{k\rightarrow\infty} \frac{\sin{kt}}{\pi t} = \delta(t),$$
meaning that
$$\int_0^\infty \frac{\sin{k t}}{\pi}\,\mathrm{d}k = \delta(t)$$
a result, admittedly, not satisfactory for mathematics stack exchange since the limit is not well defined (and is not finite) but it is noteworthy and useful nonetheless.
