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I am not sure if the following way is valid to evaluate

$$\int_{0}^{\infty}\frac{\cos(\alpha x)-\cos(\beta x)}{x}\text{d}x$$

SOURCE

I saw this and this

But I am asking if the following is valid or not?

enter image description here

Edit:

Note that $\int_{0}^{\infty}\frac{f(t)}{t}\text{d}t=\int_{0}^{\infty}\mathcal{L}\text{\{}f(t)\text{\}}\text{d}s$

$\int_{0}^{\infty}\frac{\cos(\alpha x)}{x}\text{d}x=\int_{0}^{\infty}\mathcal{L}\text{\{}\cos(\alpha x)\text{\}}\text{d}s=\int_{0}^{\infty}\frac{s}{s^2+\alpha^2}\text{d}s=\frac{1}{2}\log(s^2+\alpha^2)$, and

$\int_{0}^{\infty}\frac{\cos(\beta x)}{x}\text{d}x=\int_{0}^{\infty}\mathcal{L}\text{\{}\cos(\beta x)\text{\}}\text{d}s=\int_{0}^{\infty}\frac{s}{s^2+\beta^2}\text{d}s=\frac{1}{2}\log(s^2+\beta^2)$

Taking the difference, we get:

$\frac{1}{2}\log(\frac{s^2+\alpha^2}{s^2+\beta^2})$

Now applying the limits $s=0$ to $s \rightarrow \infty$, we get:

$\frac{1}{2}\log(1)-\frac{1}{2}\log(\frac{\alpha^2}{\beta^2})=\log(\frac{\beta}{\alpha})$

Having the right answer does not mean that the way is right.

Your help would be appreciated. Thanks!

Hussain-Alqatari
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    Please do not use pictures for critical portions of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – Martin R Sep 24 '22 at 19:10
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    @MartinR well, I will edit it now. Thanks for letting me know that pictures may not be legible. – Hussain-Alqatari Sep 24 '22 at 19:12
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    No, it is not valid to compute a convergent integral by writing as the difference of two divergent integrals. – GEdgar Sep 24 '22 at 19:43
  • @GEdgar but for some limit problems, we have (sometimes) $\infty - \infty$ and that approach a finite value, sometime. So why in this case it is not valid? Thanks. – Hussain-Alqatari Sep 24 '22 at 20:49
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    It's nearly valid. One can easily write$$\begin{align}\int_0^\infty\frac{\cos(\alpha x)-\cos(\beta x)}{x}\mathrm{d}x&=\int_0^\infty\mathcal{L}{\cos(\alpha t)-\cos(\beta t)}\mathrm{d}s\&=\int_0^\infty\left(\frac{s}{s^2+\alpha^2}-\frac{s}{s^2+\beta^2}\right)\mathrm{d}s\&=\left[\frac12\ln\frac{s^2+\alpha^2}{s^2+\beta^2}\right]_0^\infty\&=\ln\frac{\beta}{\alpha}.\end{align}$$ – J.G. Sep 24 '22 at 20:59
  • Is $$\int_0^\infty\frac{f(t)}{t};dt= \int_0^\infty\mathcal L{f(t)};ds$$valid for conditionally convergent improper integrals like this one? Reference? – GEdgar Sep 24 '22 at 21:14
  • No, it is not valid. The function $\cos{(\alpha x)}/x$ has a pole at $x =0$, so integrating from $0$ to $\infty$ results in divergence. Same with the other integral. – Accelerator Sep 24 '22 at 21:24
  • @J.G. So the my idea of using Laplace transform is valid, except I should not split the integrand. Am I right? – Hussain-Alqatari Sep 25 '22 at 05:57

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