Evaluation of the improper integral $\int_{0}^{\infty}\frac{\cos6t-\cos4t}{t}\text{d}t$ I need help calculating the following improper integral:
$$\int_{0}^{\infty}\frac{\cos6t-\cos4t}{t}\text{d}t$$
I tried using substitutions and expansions for the cosine function, but nothing worked. The answer is supposed to be $\ln(2/3)$.
 A: Try using the integral identity 
$$
\int^{\infty}_{0}\frac{f(at)-f(bt)}{t}dt=f(0)\ln(\frac{b}{a}),~a,b>0
$$
for $a=6$ and $b=4$ which gives the answer. This is called Frullani's theorem. For a proof, see the following:
Proof of Frullani's theorem
A: Our integral is just: 
$$ I=2\int_{0}^{+\infty}\frac{\sin^2(2t)-\sin^2(3t)}{t}\,dt, $$
so by exploiting:
$$ \mathcal{L}(\sin^2(2t))=\frac{8}{s(s^2+16)},\qquad  \mathcal{L}(\sin^2(3t))=\frac{18}{s(s^2+36)}$$
we have:
$$ I = -\int_{0}^{+\infty}\frac{20 s\,ds}{(s^2+16)(s^2+36)} = -\int_{0}^{+\infty}\frac{10\,du}{(u+16)(u+36)}$$
from which $I=\color{red}{\log\frac{2}{3}}$ readily follows by partial fraction decomposition.
A: EDIT:
Applying this theorem, An Extended Frullani Integral, where $f(x)=\cos(x)$, $F(x)=\int_0^x f(t)\,dt$, and $\bar F(x)=\frac1x F(x)$, we find that $F'(0)=1$ and $\lim_{x\to\infty} \bar F(x)=0$.  We find immediately that
$$\int_0^\infty \frac{\cos(6t)-\cos(4t)}{t}\,dt=\log(2/3)$$
And we are done!!


From Frullani's Theorem we have
$$\int_0^{\infty}\frac{f(ax)-f(bx)}{x}dx=(f(0)-f(\infty))\log(b/a)$$
Here, we can write
$$\begin{align}
\int_0^{\infty}\frac{\cos(6x)-\cos(4x)}{x}dx&=\text{Re}\left(\lim_{\epsilon \to 0^{+}}\int_0^{\infty}\frac{e^{(i-\epsilon)6x}-e^{(i-\epsilon)4x}}{x}dx\right)\\\\
&=\text{Re}\left(\lim_{\epsilon \to 0^{+}}\log(4/6)\right)\\\\
&=\log(2/3)
\end{align}$$

NOTE:
WE justify the interchange of the limit with the integral in the preceding development by use of Cauchy's Integral Theorem.  To that end, let $f(z)$ be the given by
$$f(z)=\frac{e^{i6z}-e^{i4z}}{z}$$
and note that $f$ is analytic with a removable discontinuity at $z=0$.  Thus, we have for any simply connected contour $C$
$$\oint_C f(z)\,dz=0$$
We choose $C$ to be the contour with the following $3$ segments
$(i)$  The line segment $C_1$ along the positive real axis from $(0,0)$ to $(R,0)$
$(ii)$  The line segment $C_2$ from $(R/\sqrt{1+\epsilon^2},R\epsilon/\sqrt{1+\epsilon^2})$ to $(0,0)$
and
$(iii)$ The circular arc $C_R$ of radius $R$ that goes from $(R,0)$ to $(R/\sqrt{1+\epsilon^2},R\epsilon/\sqrt{1+\epsilon^2})$
Then, we have
$$\begin{align}
\oint_C f(z)\,dz&=\oint_C \frac{e^{i6z}-e^{i4z}}{z}dz\\\\
&=\int_{C_1}\frac{e^{i6z}-e^{i4z}}{z}dz+\int_{C_2}\frac{e^{i6z}-e^{i4z}}{z}dz+\int_{C_R}\frac{e^{i6z}-e^{i4z}}{z}dz\\\\
&=\int_0^R \frac{e^{i6x}-e^{i4x}}{x}\,dx\\\\
&-\int_0^{R/\sqrt{1+\epsilon^2}} \frac{e^{6x(i-\epsilon)}-e^{4x(i-\epsilon)}}{x}\,dx\\\\
&+i\,\int_0^{\arctan(\epsilon)}\left(e^{i6Re^{i\theta}}-e^{i4Re^{i\theta}}\right)d\theta \tag 1\\\\
&=0
\end{align}$$
As $R\to \infty$ the last integral on the right-hand side of $(1)$ approaches $0$ as $O(R^{-1})$ by Jordan's Lemma.  Thus, we have
$$\int_0^{\infty} \frac{e^{i6x}-e^{i4x}}{x}\,dx=\int_0^{\infty} \frac{e^{6x(i-\epsilon)}-e^{4x(i-\epsilon)}}{x}\,dx$$
for all $\epsilon>0$.
A: $$I=\int_0^∞(\cos(6t)-\cos(4t))/t²dt=\int_0^∞(\cos(6t)-1)/t²dt-\int_0^∞(\cos(6t)-1)/t²dt=\\-2\int_0^∞\sin²(12t)/t²dt+2\int_0^∞\sin²(8t)/t²$$dt. Changing variables $12t->u,8t->v$ gives $$I=-8\int_0^∞sin²(t)/t²dt.$$The last integral is well-known either by residues or by Fourier's Transform.
