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How do you prove $\int^{\infty}_0e^{-ax}\frac{1-\cos(x)}{x^2}\mathrm{d}x=\arctan\frac{1}{a}-\frac{a}{2}\ln(a^2+1)$, where $a>0$?

I have no idea how to start.

koifish
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1 Answers1

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The following answer assumes a proposed correction to the problem statement.

Let $f(a):=\int_0^\infty\frac{1-\cos x}{x^2}\exp -ax dx$ so $$f''(a)=\int_0^\infty (1-\cos x)\exp -ax dx=\frac{1}{a}-\frac{a}{a^2+1}.$$Thus constants $B,\,C$ exist with $$f(a)=-\frac{a}{2}\ln (a^{-2}+1)-\arctan a+Ba+C.$$Since $f(\infty)=0\implies B=0,\,C=\frac{\pi}{2}$ and $\arctan\frac{1}{a}=\frac{\pi}{2}-\arctan a$, we're done.

J.G.
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  • Nice solution, for sure and $\to +1$ ! – Claude Leibovici Nov 06 '18 at 11:38
  • Hi, how did you get from the second equation to the third equation? ie, from f''(a) to f(a)? Did you integrate $\frac{1}{a}-\frac{a}{a^2+1}$ twice? – koifish Nov 06 '18 at 12:14
  • @YipJungHon Yes. The first integral is easy; the second is a bit trickier, unless you've ever differentiated $a\ln a$ before. However, given that we "know the answer", you can just double-check its second derivative is what we need. – J.G. Nov 06 '18 at 12:18
  • Okay, is there any particular name for this method? Don't believe I seen it before – koifish Nov 06 '18 at 12:22
  • @YipJungHon Physicists call it Feynman's trick. https://en.wikipedia.org/wiki/Leibniz_integral_rule – J.G. Nov 06 '18 at 12:51