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Compute the following integral $$\int \limits_0^{+\infty }\frac{\cos \left(\ln x\right)}{x^2+e^{2017\pi \ }}dx.$$

I came across this problem but before to start to attack it I asked myself a question: since we have infinite interval, then this is probably improper integral, right? But the lower limit of integration looks a bit weird since the integrand $\dfrac{\cos \left(\ln x\right)}{x^2+e^{2017\pi \ }}$ is not defined at $0$ and the limit does not exists at $0$? So how to understand that integral correctly?

RFZ
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    The integrand has no limit at zero, but the integral may still exist as a limit. – Gerry Myerson Jun 17 '22 at 05:20
  • @GerryMyerson, so obviously this integral is not definite one! So probably this is improper integral. But when we talk about improper we have intfinite interval OR function is unbounded in neighboorhood of some point of integration. So it does not satisfy to the definition of improper integral also. I just want to understand what is the correct definition of this integral. – RFZ Jun 17 '22 at 05:23
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    I don't know where you are getting your definitions. It is a definite integral, and it is an improper integral. $\int_0^{\infty}x^{-1/2}e^{-x},dx$ would also be an improper definite integral. – Gerry Myerson Jun 17 '22 at 05:29
  • Do you also want the closed form (if exists) for the integral? – Shimura Variety Jun 17 '22 at 06:17
  • @GerryMyerson, what is wrong with my definitions? For definite integral the function should be defined everywhere on $[0,+\infty)$. These definitions are from Zorich's book and I have never heard the notion "improper definite integral". – RFZ Jun 17 '22 at 12:41
  • I'm not familiar with Zorich's book. The only definition I've ever seen is that anything that looks like $\int_a^bf(x),dx$ is a definite integral, as opposed to an indefinite integral, which is anything that looks like $\int f(x),dx$. – Gerry Myerson Jun 17 '22 at 12:57
  • @GerryMyerson, so do we understand this original integral as $\lim\limits_{R\to \infty}\int \limits_{0}^{R}f(x)dx$ or what? – RFZ Jun 17 '22 at 14:55
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    https://math.stackexchange.com/questions/837556/how-to-compute-this-integral-involving-sech?noredirect=1 or https://math.stackexchange.com/questions/804483/do-these-integrals-have-a-closed-form-i-1-int-infty-infty-frac-s?noredirect=1 – Svyatoslav Jun 17 '22 at 17:37
  • I'd say, $\lim_{a\to0}\lim_{b\to\infty}\int_a^bf(x),dx$. – Gerry Myerson Jun 17 '22 at 23:41
  • @GerryMyerson, hmm. I have never seen like this before to be honest. 1) is that definition well-defined? I mean what if we change the order of limits? – RFZ Jun 18 '22 at 00:31
  • I don't think the order of the limits makes any difference in this case. – Gerry Myerson Jun 18 '22 at 01:36
  • The Moore-Osgood Theorem might apply, https://en.wikipedia.org/wiki/Iterated_limit#Sufficient_condition – Gerry Myerson Jun 18 '22 at 01:43
  • @GerryMyerson, Thank you for the link! What about the following approach? We can define the integrand at $0$ arbitrarily, then it is integrable on $[0,R]$ for each $R$. Then one can consider this integral as a $\lim_{R\to \infty} \int_{0}^{R}f(x)dx$. Is that correct? – RFZ Jun 18 '22 at 03:14
  • Yes, looks like that will work. – Gerry Myerson Jun 18 '22 at 03:15
  • Another way: pick a number $t$, $0<t<\infty$, and write $\int_0^{\infty}f(x),dx=\int_0^tf(x),dx+\int_t^{\infty}f(x),dx=\lim_{a\to0}\int_a^tf(x),dx+\lim_{b\to\infty}\int_t^bf(x),dx$, a sum of two simpler improper integrals. – Gerry Myerson Jun 19 '22 at 23:55
  • @GerryMyerson, I am really confused. I don't think that you can consider $\int_0^t f(x)dx$ as an improper integral because in the definition of improper integral either the interval should be infinite or the interval is finite but the function is unbounded in one of the endpoints. This definition was from Zorich where I learnt mathematical analsis and the same definition in wikipedia. – RFZ Jun 20 '22 at 00:25
  • I'd consider it to be improper because the integrand is not defined at $x=0$, but I accept that there is merit in the definitions you cite. – Gerry Myerson Jun 20 '22 at 03:16

2 Answers2

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Surprizing or not, a CAS gives (for $k>0$) $$\int \frac{\cos (\log (x))}{x^2+k}\,dx=\frac{(1-i) x^{1+i} \, _2F_1\left(\frac{1+i}{2},1;\frac{3+i}{2};-\frac{x^2}{k}\right) +(1+i) x^{1-i} \, _2F_1\left(\frac{1-i}{2},1;\frac{3-i}{2};-\frac{x^2}{k}\right) }{4 k}$$ where appear two Gaussian hypergeometric function.

For an infinite value of $x$, it is $$\frac{1}{4} \pi \left(1+k^i\right) k^{-\frac{1+i}{2}} \text{sech}\left(\frac{\pi }{2}\right)$$

In fact, around $x=0$ there is no problem since $$\frac{\cos (\log (x))}{x^2+k}=\left(\frac{1}{k}+O\left(x^2\right)\right) \cos (\log (x))$$ $$\int \cos (\log (x))\,dx=\frac{1}{2} x (\sin (\log (x))+\cos (\log (x))) \quad \to \quad 0$$

Claude Leibovici
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The integrand (defined arbitrarily at $x=0$) is Riemann integrable on $[0,R]$ for any finite $R$, as it is bounded and almost everywhere continuous. So strictly speaking you don't need to consider it as improper at $0$. That does not prevent you from treating it as improper there, i.e. it's still true that $$ \int_0^\infty f(x)\; dx = \lim_{R \to \infty} \lim_{\epsilon \to 0+} \int_\epsilon^R f(x)\; dx$$ For example, you might use this with some numerical method that works well on $[\epsilon, R]$ but not on $(0, R)$.

Robert Israel
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  • So you can consider this integral as the double limit as you have shown and of course as $\lim \limits_{R\to \infty}\int \limits_{0}^{R}f(x)dx$, right? – RFZ Jun 17 '22 at 12:46