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While seeing this post, the following integral is just struck me

\begin{equation} \int_0^\infty \frac{dx}{(1+x^2)(1+\tan x)}\tag1 \end{equation}

I have tried like what user @OlivierOloa did in his answer, but no luck with finding its closed-form. Using substitution $x\mapsto\tan x$ didn't make it any easier either.

\begin{equation} \int_0^{\pi/2}\frac{dx}{1+\tan (\tan x)}\tag2 \end{equation}

I reached a dead end in the attempt. How does one evaluate the integral $(1)$ or $(2)$?

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Sophie Agnesi
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    The integral (1) diverges, since cosine has a lot of positive zeros. – mickep Jun 09 '16 at 04:29
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    I failed to see why this OP got downvoted since it's a fair question that posted according to the terms and conditions of MSE. @downvoter: Care to explain? – Sophie Agnesi Jun 09 '16 at 04:43
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    @SophieAgnesi Cannot speak for that person, but a wild guess would be that it's because your integral blatantly diverges: the denominator cancels at every $k\pi+\frac{\pi}{2}$. (Same thing for the integrand of (2), incidentally, at some value $\approx 1$ for instance.) – Clement C. Jun 09 '16 at 04:48
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    @ClementC. Thanks for your comment. Maybe it diverges, so if one posts a divergent integral, I presume I can downvote it, too? – Sophie Agnesi Jun 09 '16 at 04:53
  • To be blunt, nothing can prevent you from doing that. To be even blunter, it'd be in my opinion quite dumb, the same way that this downvote seems rather excessive and unjustified to me. Then, again, the person that did that may have had a good reason, and chose not to share it. – Clement C. Jun 09 '16 at 04:56
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    @ClementC. I'm just annoyed with this anonymous downvote, but your comment is fair enough. – Sophie Agnesi Jun 09 '16 at 05:14
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    On the contrary, I think that working on rather complicated integrals, one should be able to spot obvious divergence. – mickep Jun 09 '16 at 05:14
  • @mickep I still believe that this integral doesn't diverge. The integrand itself oscillates infinitely often inside $\left[0,\frac{\pi}{2}\right]$. – Sophie Agnesi Jun 09 '16 at 07:46
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    If we look at (1), the $1/\cos x$ part goes like $1/(\pi/2-x)+O(\pi/2-x)$ close to $x=\pi/2$. That singularity is non-integrable. You have the same problem at $x=\pi/2+k\pi$ for each positive integer $k$. Since (2) is just a change of variables from (1) it has the same problem (but in other points). – mickep Jun 09 '16 at 08:20
  • Osciliating inf often does not always imply convergence. – Math-fun Jun 09 '16 at 11:31
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    Oscillation is not related to the divergence of the integral. The integral does not exist simply because the function $1/(x^2+1)(1+\tan x)$ develops simple poles at $x = \frac{3\pi}{4} + k\pi$ for each $k \in \mathbb{Z}$. So the integral does not make sense either in improper Riemann integral sense or in Lebesgue integral sense. It does makes sense in Cauchy principal value sense with the value approximately $$ 0.79198998090512221622\cdots, $$ but within my knowledge, it seems hopeless to obtain a closed form for this. – Sangchul Lee Mar 05 '19 at 05:11
  • Now there are many alternative answers. I have tried to obtain understable answer there. I hope, that it was successful attempt. – Yuri Negometyanov Mar 08 '23 at 23:11

1 Answers1

1

$$\int_0^{\infty } \frac{1}{\left(1+x^2\right) (1+\tan (x))} \, dx=\\\int_0^{\infty } \left(\mathcal{L}_x\left[\frac{1}{1+\tan (x)}\right](t)\right) \left(\mathcal{L}_x^{-1}\left[\frac{1}{1+x^2}\right](t)\right) \, dt=\Re\left(\int_0^{\infty } \frac{\left((1+i)+e^{\frac{\pi t}{4}} t B_i\left(\frac{i t}{2},0\right)\right) \sin (t)}{2 t} \, dt\right)=\\\frac{\pi }{4}+\int_0^{\infty } \frac{1}{2} e^{\frac{\pi t}{4}} \Re\left(B_i\left(\frac{i t}{2},0\right)\right) \sin (t) \, dt=\\\frac{\pi }{4}+\sum _{k=1}^{\infty } \left(\frac{1}{2} e^{-2 k} \pi \sin \left(\frac{k \pi }{2}\right)+\cos \left(\frac{k \pi }{2}\right) (-\text{Chi}(2 k) \sinh (2 k)+\cosh (2 k) \text{Shi}(2 k))\right)=\\\frac{\pi }{4}+\frac{e^2 \pi }{2 \left(1+e^4\right)}+\sum _{k=1}^{\infty } \left(\cos \left(\frac{k \pi }{2}\right) \cosh (2 k) \text{Shi}(2 k)-\cos \left(\frac{k \pi }{2}\right) \text{Chi}(2 k) \sinh (2 k)\right)=\\\frac{\pi }{4}+\frac{e^2 \pi }{2 \left(1+e^4\right)}+\sum _{k=1}^{\infty } \left(-\frac{1}{2} e^{2 k} \cos \left(\frac{k \pi }{2}\right) \text{Ei}(-2 k)+\frac{1}{2} e^{-2 k} \cos \left(\frac{k \pi }{2}\right) \text{Ei}(2 k)\right)=\\\color{blue}{-\int_0^1 \left(\frac{e^4}{\left(1+e^4\right) x}-\frac{e^4}{\left(e^4+e^{4 x}\right) x}\right) \, dx+M}$$

where:$B_i$ is incomplete beta function,$\text{Chi}$ is hyperbolic cosine integral function ,$\text{Shi}$ is hyperbolic sine integral function, $\text{Ei}$ is the exponential integral function.

where: $\color{blue}{M}$ -(Mathematica code) is:

(2*(-1 + E^4)*EulerGamma + (1 + E^2)^2*Pi - Log[4] + I*E^6*Log[4] + E^4*Log[16] - Derivative[1, 0][PolyLog][0, (-I)/E^2] - E^4*Derivative[1, 0][PolyLog][0, (-I)/E^2] - Derivative[1, 0][PolyLog][0, I/E^2] - E^4*Derivative[1, 0][PolyLog][0, I/E^2] + Derivative[1, 0][PolyLog][0, (-I)*E^2] + E^4*Derivative[1, 0][PolyLog][0, (-I)*E^2] + Derivative[1, 0][PolyLog][0, E^8] + E^4*Derivative[1, 0][PolyLog][0, E^8] - E^4*Derivative[0, 1, 0][HurwitzLerchPhi][E^8, 0, 1/2] - E^8*Derivative[0, 1, 0][HurwitzLerchPhi][E^8, 0, 1/2] - I*E^6*Derivative[0, 1, 0][HurwitzLerchPhi][E^8, 0, 3/4] - I*E^10*Derivative[0, 1, 0][HurwitzLerchPhi][E^8, 0, 3/4] + I*E^10*Derivative[0, 1, 0][HurwitzLerchPhi][E^8, 0, 5/4] + I*E^14*Derivative[0, 1, 0][HurwitzLerchPhi][E^8, 0, 5/4])/(4*(1 + E^4))

Mariusz Iwaniuk
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