Evaluation of the improper integral $\int_{0}^{\infty}\frac{\cos6t-\cos4t}{t}\text{d}t$

Question

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 $\ln(2/3)$.

Answer 1

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

Answer 2

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.

Answer 3

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!!

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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

$$\begin{align} \int_0^{\infty}\frac{\cos(6x)-\cos(4x)}{x}dx&=\text{Re}\left(\lim_{\epsilon \to 0^{+}}\int_0^{\infty}\frac{e^{(i-\epsilon)6x}-e^{(i-\epsilon)4x}}{x}dx\right)\\\\ &=\text{Re}\left(\lim_{\epsilon \to 0^{+}}\log(4/6)\right)\\\\ &=\log(2/3) \end{align}$$

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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 $\epsilon>0$.

Answer 4

$$I=\int_0^∞(\cos(6t)-\cos(4t))/t²dt=\int_0^∞(\cos(6t)-1)/t²dt-\int_0^∞(\cos(6t)-1)/t²dt=\\-2\int_0^∞\sin²(12t)/t²dt+2\int_0^∞\sin²(8t)/t²$$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.