improper integrals - Evaluating $int_0^{infty} frac{sin xt sin yt cos zt}{t^2} , dt$

The problem is to evaluate the improper integral $I = \int_0^{\infty} \frac{\sin xt \sin yt \cos zt}{t^2} dt$.

This can be written as $\int_0^{\infty} dt \int_0^y \frac{\sin xt \cos st \cos zt}{t} ds$, noting that $\int_0^y \cos(st) ds = \frac{\sin yt}{t}$.

I want to interchange the order of the two integrals for $I$ show that:

$$I = \int_0^{\infty} dt \int_0^y \frac{\sin xt \cos st \cos zt}{t} ds = \int_0^y ds \int_0^{\infty} \frac{\sin xt \cos st \cos zt}{t} dt$$

I know how to evaluate the second integral. The integrand can be rewritten using the product to sum trigonometric identities to get:

$$f(s) = \int_0^{\infty} \frac{\sin xt \cos st \cos zt}{t} dt =
\frac14 \int_0^{\infty} \frac{\sin((x+||s|-|z||)t)}{t} + \frac{\sin((x-||s|-|z||)t)}{t} +$$
$$\frac{\sin((x+||s|+|z||)t)}{t} + \frac{\sin((x-||s|+|z||)t)}{t} dt$$

It is well known that $ \int_0^{\infty} \frac{\sin xt}{t} dt =\frac{\pi}{2}$ sgn $x$, so we can provide an explicit expression for the integral.

In order to interchange the integrals for $I$, don't I have to show that $f(s) = \int_0^{\infty} \frac{\sin xt \cos st \cos zt}{t} dt$ converges uniformly on the interval {$0 \le s \le y$}. But $ g(x) = \int_0^{\infty} \frac{\sin xt}{t} dt$ does not converge uniformly on any interval that contains $0$ and $f(s)$ contains four integrals of this form.

So I don't know how to show that I can interchange the order of the integrals or if it even valid to do so.

Answer

We can use the addition formulas for $\sin$ and $\cos$, or more easily the relations $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$ and $\cos(z) = \frac{e^{iz}+e^{-iz}}{2}$, to show that

$$\matrix{4\sin(xt)\sin(yt)\cos(zt) &=& -\cos((x+y+z)t) + \cos((x-y-z)t) \\&&- \cos((x+y-z)t)+\cos((x-y+z)t)}$$

Using this the integral we are after can be written in terms of

$$I(w) \equiv \int_0^\infty \frac{\cos(wt)-1}{t^2}{\rm d}t = |w|\int_0^\infty \frac{\cos(t)-1}{t^2}{\rm d}t = -\frac{\pi}{2}|w|$$

which is derived below as

$$\int_0^\infty\frac{\sin(xt)\sin(yt)\cos(zt)}{t^2}{\rm d}t = \frac{1}{4}\left[-I(x+y+z) + I(x-y-z) - I(x+y-z) + I(x-y+z)\right]\\ = \frac{\pi}{8}\left[\left| x+y+z\right|-\left| x-y-z\right|+\left| x+y-z\right| -\left| x-y+z\right| \right]$$

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To compute the integral $\int_0^\infty \frac{\cos(t)-1}{t^2}{\rm d}t$ we generalize it by adding a $e^{-wt}$ term to the integrand, i.e. we study

$$f(w) = \int_0^\infty \frac{\cos(t)-1}{t^2}e^{-wt}{\rm d}t$$

We now expand $\cos(t)$ in a Taylor-series to get

$$f(w) = \sum_{k=1}^\infty \frac{(-1)^k}{(2k)!}\int_0^\infty t^{2k-2}e^{-wt}{\rm d}t = w\sum_{k=1}^\infty \frac{(-1/w^2)^k}{(2k)(2k-1)}$$

where I have used the definition of the $\Gamma$-function to perform the middle integral. The justification for exchanging the summation and integration can be found in this answer. Splitting $\frac{1}{2k(2k-1)}$ into partial fractions we get

$$f(w) = w\sum_{k=1}^\infty \frac{(-1/w^2)^k}{2k-1} - w\sum_{k=1}^\infty \frac{(-1/w^2)^k}{2k} = -w\arctan\left(\frac{1}{w}\right) + \frac{1}{2}\log\left(1 +\frac{1}{w^2}\right)$$

where we have used the Taylor-series for $\log(1+x)$ and $\arctan(x)$ to evaluate the sums. Taking the limit $w\to 0$ gives us the desired result $-\frac{\pi}{2}$ (using e.g. L'Hopitals).