Evaluate $\int\limits_{0}^{+\infty} \frac{1-\cos(ax)}{x}\cos(bx)\operatorname d\!x ,\ \ a > 0, \ b > 0$

Question

How can I evaluate this integral?

$$\int\limits_{0}^{+\infty} \frac{1-\cos(ax)}{x}\cos(bx)\operatorname d\!x ,\ \ a > 0, \ b > 0$$

It looks like it can be transformed into Frullani's integral via subsitution but I have no idea how to do it.

Wolfram Alpha evaluates it as:

$$\frac{1}{4}\ln\frac{(a^2-b^2)^2}{b^4}$$

Answer 1

Using the identity $$\cos ax\cos bx=\frac{1}{2}(\cos(a+b)x+\cos(a−b)x),$$ the integral can be written $$\frac{1}{2}\int\limits_{0}^\infty \frac{\cos(bx)-\cos(a+b)x}{x}\,dx+\frac{1}{2}\int\limits_{0}^\infty \frac{\cos(bx)-\cos(a-b)x}{x}\,dx=\frac{1}{2}\ln\frac{a+b}{b}+\frac{1}{2}\ln\frac{a-b}{b}.$$ One version of the conditions for Frullani integrals is satisfied: $\lim_{x\rightarrow0+}\cos x$ exists ($=1$), and the integrals $\int^{X_2}_{X1}\cos x\,dx$ are bounded.

Answer 2

Well, we can look at the Laplace transform of the function:

$$\text{f}\left(x\right):=\frac{1-\cos\left(\text{a}x\right)}{x}\cdot\cos\left(\text{b}x\right)\tag1$$

We get:

$$\text{F}\left(\text{s}\right):=\mathscr{L}_x\left[\frac{1-\cos\left(\text{a}x\right)}{x}\cdot\cos\left(\text{b}x\right)\right]_{\left(\text{s}\right)}:=\int_0^\infty\frac{1-\cos\left(\text{a}x\right)}{x}\cdot\cos\left(\text{b}x\right)\cdot e^{-\text{s}x}\space\text{d}x\tag2$$

Now, using the 'frequency-domain integration' property of the laplace transform:

$$\text{F}\left(\text{s}\right)=\int_\text{s}^\infty\mathscr{L}_x\left[\left(1-\cos\left(\text{a}x\right)\right)\cdot\cos\left(\text{b}x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma=$$ $$\int_\text{s}^\infty\left(\mathscr{L}_x\left[\cos\left(\text{b}x\right)\right]_{\left(\sigma\right)}-\mathscr{L}_x\left[\cos\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)\right]_{\left(\sigma\right)}\right)\space\text{d}\sigma\tag3$$

Use the Laplace transform of the cosine function:

$$\text{F}\left(\text{s}\right)=\int_\text{s}^\infty\left(\frac{\sigma}{\sigma^2+\text{b}^2}-\mathscr{L}_x\left[\cos\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)\right]_{\left(\sigma\right)}\right)\space\text{d}\sigma\tag4$$

Use:

$$\cos\left(\text{a}x\right)\cdot\cos\left(\text{b}x\right)=\frac{\cos\left(\left(\text{a}-\text{b}\right)x\right)+\cos\left(\left(\text{a}+\text{b}\right)x\right)}{2}\tag5$$

So, for the Laplace transform we end up with:

$$\text{F}\left(\text{s}\right)=\int_\text{s}^\infty\left(\frac{\sigma}{\sigma^2+\text{b}^2}-\frac{1}{2}\cdot\left(\frac{\sigma}{\sigma^2+\left(\text{a}-\text{b}\right)^2}+\frac{\sigma}{\sigma^2+\left(\text{a}+\text{b}\right)^2}\right)\right)\space\text{d}\sigma\tag6$$

Find: $\text{s}=0$.

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For the indefinite general integral we know:

$$\text{y}\left(x,\text{n}\right):=\int\frac{x}{x^2+\text{n}^2}\space\text{d}x=\frac{\ln\left|x^2+\text{n}^2\right|}{2}+\text{C}\tag7$$

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So, for $\text{F}\left(\text{s}\right)$ at $\text{s}=0$ we get:

$$\text{F}\left(0\right)=\lim_{\text{p}\to\infty}\space\left[\frac{\ln\left|\sigma^2+\text{b}^2\right|}{2}-\frac{1}{2}\cdot\left(\frac{\ln\left|\sigma^2+\left(\text{a}-\text{b}\right)^2\right|}{2}+\frac{\ln\left|\sigma^2+\left(\text{a}+\text{b}\right)^2\right|}{2}\right)\right]_0^\text{p}\tag8$$

• When $\sigma=0$: $$\frac{\ln\left|0^2+\text{b}^2\right|}{2}-\frac{1}{2}\cdot\left(\frac{\ln\left|0^2+\left(\text{a}-\text{b}\right)^2\right|}{2}+\frac{\ln\left|0^2+\left(\text{a}+\text{b}\right)^2\right|}{2}\right)=$$ $$\frac{\ln\left|\text{b}^2\right|}{2}-\frac{1}{2}\cdot\left(\frac{\ln\left|\left(\text{a}-\text{b}\right)^2\right|}{2}+\frac{\ln\left|\left(\text{a}+\text{b}\right)^2\right|}{2}\right)\tag9$$

  • Try to prove that: $$\lim_{\text{p}\to\infty}\space\left(\frac{\ln\left|\text{p}^2+\text{b}^2\right|}{2}-\frac{1}{2}\cdot\left(\frac{\ln\left|\text{p}^2+\left(\text{a}-\text{b}\right)^2\right|}{2}+\frac{\ln\left|\text{p}^2+\left(\text{a}+\text{b}\right)^2\right|}{2}\right)\right)=0\tag{10}$$

End, so we end up with:

$$\text{F}\left(0\right)=-\left(\frac{\ln\left|\text{b}^2\right|}{2}-\frac{1}{2}\cdot\left(\frac{\ln\left|\left(\text{a}-\text{b}\right)^2\right|}{2}+\frac{\ln\left|\left(\text{a}+\text{b}\right)^2\right|}{2}\right)\right)=\ln\left(\sqrt{\left|\frac{\text{a}^2}{\text{b}^2}-1\right|}\right)\tag{11}$$