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As title, how do I evaluate the integral $\int_0^{\infty} \frac{\cos(\pi x)-\cos(e x)}{x} dx$? My answer key provides the answer $\ln(\frac{e}{\pi})$, but I don't understand how at all. I have tried infinite series, differentiation under integration sign and integration by parts, but either I am doing it incorrectly or the integral always diverges. Any hints?

Thank you so much,

Gareth

Gareth Ma
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    See Frullani's integral – user170231 Oct 14 '21 at 17:21
  • @user170231 $f(x)=\cos(x)$ does not have a limit at $\infty$. – Gareth Ma Oct 14 '21 at 17:22
  • @GarethMa Squeeze theorem produces a limit of $0$. – UNOwen Oct 14 '21 at 17:23
  • @UNOwen On the wikipedia page linked above it says that $f(\infty)$ is required to exist but $\cos(x)$ does not converge, not $\frac{f(x)}{x}$ (which I know is $0$). – Gareth Ma Oct 14 '21 at 17:24
  • Yes, you are correct. See https://www.degruyter.com/document/doi/10.1515/math-2017-0001/html, which has your specific case at Example $3.5$. Using a series expansion does the trick. – UNOwen Oct 14 '21 at 17:30
  • @Gareth Ma Frullani's integral result holds for weaker assumptions: see this answer on linked question with $f(x) = 1 - \cos x$ – K B Dave Oct 14 '21 at 17:33
  • So, the general formula is if $f(x)=\sum_{0}^{\infty} \phi_k C(k) x^{\alpha k}$, then your integral is $C(0)\cdot \ln\left(\frac{b}{a}\right)$. Proof is provided after Theorem 3.1. Hope this helps. – UNOwen Oct 14 '21 at 17:34
  • @KBDave Yes, and it is fairly trivial to find $A=1$ and $B=0$ in this case. – UNOwen Oct 14 '21 at 17:38
  • @UNOwen to be honest the problem was for a integration bee (practice), so I thought there must be a simpler way than using some advanced theorem (which is out of syllabus). Thanks anyways, I’ve posted my answer. Sad that it got marked as duplicate – Gareth Ma Oct 15 '21 at 07:50

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Nevermind, I have solved my own problem without the use of advanced theorems (such as Frullani's theorem).

Let $I(t)=\int_0^{\infty} e^{-tx}\cdot\frac{\cos(ax)}{x}dx$. Differentiating gives $I'(t)=-\int_0^{\infty} e^{-tx}\cos(ax)dx$ which can then be evaluated by integrating by parts (twice). Finally we get $I'(t)=-\frac{t}{a^2+t^2}$ i.e. $I(t)=-\frac{1}{2}\ln(a^2+t^2)$. Evaluating at $t=0$ gives $\int_0^{\infty}\frac{\cos(ax)}{x}dx=-\ln a$, and the final answer is $\ln\frac{e}{\pi}$.

Gareth Ma
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