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$$\DeclareMathOperator{\sech}{sech} \int \frac{- (\arctan{ \frac{(a + \sqrt{a})x}{a-1}}) \sech^2(\pi x)(-2 \pi x + sinh(2 \pi x)) \tanh^2{\pi x}}{x^4}dx =???? $$

Alternatively: $$\int_{-\infty}^{\infty} \frac{- (\arctan{ \frac{(a + \sqrt{a})x}{a-1}}) \sech^2(\pi x)(-2 \pi x + sinh(2 \pi x)) \tanh^2{\pi x}}{x^4}dx =???? $$

For some $\forall a \in \mathbb R_{\gt 1}$

This integral came up in my research and I thought, "Oh I'll just run it through Wolfram Alpha". That did not work. But now I'm greatly confused as to why. I posted this question previously, but it was closed because individuals needed some additional information, which I've provided below.

I will settle for a calculation in the range of integration from $-\infty$ to $\infty$, but I really am asking about the anti-derivative.

I was going to let this go, but now I have to know why this is so difficult to calculate in a reasonable amount of time. Anyone know?

(Naive) Motiviation:

Typically (that is to say in my experience) these types of integrals of trigonometric products can be broken down using existing methods; We can usually reach some "solved point" with some additive and/or multiplicative constants notwithstanding. "What is so special about this particular function?" is another way to phrase what I'm asking.

Some background on preferences and pitfalls when dealing with trigonometric integrals:

Showing $\int_0^{\int_0^u{\rm sech}vdv}\sec vdv\equiv u$ and $\int_0^{\int_0^u\sec vdv}{\rm sech} vdv\equiv u$ In trig substitutions, why favor $\sin$, $\tan$, $\sec$, $\sinh$, $\cosh$, $\tanh$ over $\cos$, $\cot$, $\csc$, etc? https://en.wikipedia.org/wiki/Integral_of_secant_cubed

EDIT: I changed the scalar argument of the the arctan function, so calculating the definite integral is not so obvious. Thanks to @KevinDietrich for the catch.

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    $\begingroup$ If you write down an integral like this, chance are its indefinite integral is not expressed in finite terms involving previously-named functions. $\endgroup$
    – GEdgar
    Commented Aug 2, 2023 at 14:18
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    $\begingroup$ @TymaGaidash I address the reason for closure (which apparently was done by malicious individuals and StackExchange reversed) in my question. $\endgroup$ Commented Aug 2, 2023 at 15:27
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    $\begingroup$ @GEdgar Please enlighten us! What makes you suspect something like this won't have an indefinite integral? My initial suspicions went the other way. $\endgroup$ Commented Aug 2, 2023 at 15:28
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    $\begingroup$ @KevinDietrich You are right, actually I need to edit the question so that the scalar in the $arctan$ function is variable $\in \mathbb R$. $\endgroup$ Commented Aug 2, 2023 at 15:29
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    $\begingroup$ After some strenuous calculation, the definite integral equals to$$ I\left( t \right) =\left( \frac{248}{\pi ^2}\zeta \left( 5 \right) -\frac{28}{3}\zeta \left( 3 \right) \right) t+\left( \frac{28}{3\pi ^2}\zeta \left( 3 \right) -\frac{8}{3}\ln \left( 2 \right) -\frac{4}{3}\gamma \right) t^3-\frac{4}{3}t^3\psi ^{\left( 0 \right)}\left( \frac{1}{2}+\frac{1}{t} \right) -\frac{t^5}{3}\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{t^2x^2+\pi ^2}}\mathrm{d}x+\frac{2}{3}t^5\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{\left( t^2x^2+\pi ^2 \right) ^2}}\mathrm{d}x $$ $\endgroup$
    – oO_ƲRF_Oo
    Commented Aug 5, 2023 at 9:11

1 Answer 1

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To solve this integral, I shall establish few facts: \begin{align*} &I_1=\int_0^{\infty}{\frac{\tanh ^3\left( x \right)}{x^3}}\mathrm{d}x=\frac{186}{\pi ^4}\zeta \left( 5 \right) -\frac{7}{\pi ^2}\zeta \left( 3 \right) \\&I_2=\int_0^{\infty}{\frac{\tanh ^2\left( x \right)}{x^2}}\mathrm{d}x=\frac{14}{\pi ^2}\zeta \left( 3 \right) \\&I_3=\int_{-\infty}^{\infty}{\frac{\tanh \left( x \right)}{x\left( t^2x^2+\pi ^2 \right)}}\mathrm{d}x=\frac{2}{\pi ^2}\psi ^{\left( 0 \right)}\left( \frac{1}{2}+\frac{1}{t} \right) +\frac{4}{\pi ^2}\ln \left( 2 \right) +\frac{2\gamma}{\pi ^2} \\&I_4=\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{t^2x^2+\pi ^2}}\mathrm{d}x=\frac{1}{t}-\frac{2}{t\pi ^2}\psi ^{\left( 1 \right)}\left( \frac{1}{2}+\frac{1}{t} \right) \\&I_5=\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{\left( t^2x^2+\pi ^2 \right) ^2}}\mathrm{d}x=\frac{1}{\pi ^2}\left( \frac{1}{t^2\pi ^2}\psi ^{\left( 2 \right)}\left( \frac{1}{2}+\frac{1}{t} \right) -\frac{1}{t\pi ^2}\psi ^{\left( 1 \right)}\left( \frac{1}{2}+\frac{1}{t} \right) +\frac{1}{2t} \right) \end{align*} All results above works for $t>1$
The derivation of the first two integrals one can be found here.
To calculate the third and the fourth integral, you integrate them along an upper-semicircle contour and apply residue theorem. You should get a quite messy infinite series and a tangent term, after some simplification and using Mittag-leffler series, you would arrive at the result atop.
Lastly, we have $I_5=\frac{1}{\pi ^2}\left( I_4+\frac{t}{2}\frac{\mathrm{d}}{\mathrm{d}t}I_4 \right)$. So everything justified.
Now the definite integral at hand (the indefinite one is too hard for me, it just looks undoable for me) \begin{align*} I\left( a \right) &=\int_{-\infty}^{\infty}{\frac{\sinh \left( 2\pi x \right) -2\pi x}{x^4}}\mathrm{arctan} \left( ax \right) \tanh ^2\left( \pi x \right) \mathrm{sech} ^2\left( \pi x \right) \mathrm{d}x \\& =2\int_{-\infty}^{\infty}{\frac{\sinh \left( \pi x \right) \cosh \left( \pi x \right) -\pi x}{x^4}}\mathrm{arctan} \left( ax \right) \tanh ^2\left( \pi x \right) \mathrm{sech} ^2\left( \pi x \right) \mathrm{d}x \\& =2\int_{-\infty}^{\infty}{\frac{\mathrm{arctan} \left( ax \right)}{x^4}\tanh ^3\left( \pi x \right)}\mathrm{d}x-2\pi \int_{-\infty}^{\infty}{\frac{\mathrm{arctan} \left( ax \right)}{x^3}}\tanh ^2\left( \pi x \right) \mathrm{sech} ^2\left( \pi x \right) \mathrm{d}x \\& \stackrel{\pi x\rightsquigarrow x}{=}2\pi ^3\left( \int_{-\infty}^{\infty}{\frac{\mathrm{arctan} \left( \frac{ax}{\pi} \right)}{x^4}\tanh ^3\left( x \right)}\mathrm{d}x-\int_{-\infty}^{\infty}{\frac{\mathrm{arctan} \left( \frac{ax}{\pi} \right)}{x^3}}\tanh ^2\left( x \right) \mathrm{sech} ^2\left( x \right) \mathrm{d}x \right) \\& \stackrel{\mathrm{IBP}}{=}\frac{2}{3}\pi ^4a\int_{-\infty}^{\infty}{\frac{\tanh ^3\left( x \right)}{x^3\left( a^2x^2+\pi ^2 \right)}}\mathrm{d}x \\& =\frac{4}{3}\pi ^2a\int_0^{\infty}{\frac{\tanh ^3\left( x \right)}{x^3}}\mathrm{d}x-\frac{2}{3}a^3\pi ^2\int_0^{\infty}{\frac{\tanh ^3\left( x \right)}{x\left( a^2x^2+\pi ^2 \right)}}\mathrm{d}x \\& =\frac{4}{3}\pi ^2a\int_0^{\infty}{\frac{\tanh ^3\left( x \right)}{x^3}}\mathrm{d}x-\frac{2}{3}a^3\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh \left( x \right)}{x\left( a^2x^2+\pi ^2 \right)}}\mathrm{d}x+\frac{2}{3}a^3\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh \left( x \right) \mathrm{sech} ^2\left( x \right)}{x\left( a^2x^2+\pi ^2 \right)}}\mathrm{d}x \\& \stackrel{\mathrm{IBP}}{=}\frac{4}{3}\pi ^2a\int_0^{\infty}{\frac{\tanh ^3\left( x \right)}{x^3}}\mathrm{d}x-\frac{2}{3}a^3\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh \left( x \right)}{x\left( a^2x^2+\pi ^2 \right)}}\mathrm{d}x+\frac{a^3}{3}\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{x^2}}\mathrm{d}x \\& \,\,\,\,\,\,\,\,-\frac{a^5}{3}\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{a^2x^2+\pi ^2}}\mathrm{d}x+\frac{2}{3}a^5\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{\left( a^2x^2+\pi ^2 \right) ^2}}\mathrm{d}x \\& =\frac{4}{3}\pi ^2a\int_0^{\infty}{\frac{\tanh ^3\left( x \right)}{x^3}}\mathrm{d}x+\frac{2}{3}a^3\int_0^{\infty}{\frac{\tanh ^2\left( x \right)}{x^2}}\mathrm{d}x-\frac{2}{3}a^3\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh \left( x \right)}{x\left( a^2x^2+\pi ^2 \right)}}\mathrm{d}x \\& \,\,\,\,\,\,\,\,-\frac{a^5}{3}\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{a^2x^2+\pi ^2}}\mathrm{d}x+\frac{2}{3}a^5\pi ^2\int_{-\infty}^{\infty}{\frac{\tanh ^2\left( x \right)}{\left( a^2x^2+\pi ^2 \right) ^2}}\mathrm{d}x \\& =\frac{4}{3}\pi ^2aI_1+\frac{2}{3}a^3I_2-\frac{2}{3}a^3\pi ^2I_3-\frac{a^5}{3}I_4+\frac{2}{3}a^5\pi ^2I_5 \end{align*} Pulgging in the value, we have $$ I\left( a \right) =\left( \frac{248}{\pi ^2}\zeta \left( 5 \right) -\frac{28}{3}\zeta \left( 3 \right) \right) a+\left( \frac{28}{3\pi ^2}\zeta \left( 3 \right) -\frac{8}{3}\ln \left( 2 \right) -\frac{4}{3}\gamma -\frac{4}{3}\psi ^{\left( 0 \right)}\left( \frac{1}{2}+\frac{1}{a} \right) +\frac{2}{3\pi ^2}\psi ^{\left( 2 \right)}\left( \frac{1}{2}+\frac{1}{a} \right) \right) a^3 $$ To put it in a form verbatim to your question $$ -\int_{-\infty}^{\infty}{\frac{\sinh \left( 2\pi x \right) -2\pi x}{x^4}}\mathrm{arctan} \left( \frac{a+\sqrt{a}}{a-1}x \right) \tanh ^2\left( \pi x \right) \mathrm{sech} ^2\left( \pi x \right) \mathrm{d}x \\ =-\left( \frac{248}{\pi ^2}\zeta \left( 5 \right) -\frac{28}{3}\zeta \left( 3 \right) \right) \frac{a+\sqrt{a}}{a-1}-\left( \frac{28}{3\pi ^2}\zeta \left( 3 \right) -\frac{8}{3}\ln \left( 2 \right) -\frac{4}{3}\gamma -\frac{4}{3}\psi ^{\left( 0 \right)}\left( \frac{1}{2}+\frac{a-1}{a+\sqrt{a}} \right) +\frac{2}{3\pi ^2}\psi ^{\left( 2 \right)}\left( \frac{1}{2}+\frac{a-1}{a+\sqrt{a}} \right) \right) \left( \frac{a+\sqrt{a}}{a-1} \right) ^3 $$ The numerical value check

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  • $\begingroup$ Kudos! I have part two if you're up for it: math.stackexchange.com/q/4748576/322359 I've got to do some more work to check the Desmos data. $\endgroup$ Commented Aug 6, 2023 at 18:02
  • $\begingroup$ @user3108815 I will take a look, but no promises. This one is already on my limits. $\endgroup$
    – oO_ƲRF_Oo
    Commented Aug 8, 2023 at 7:30
  • $\begingroup$ Any idea so far? I suspect something like Ramanujan's master theorem and some of what you've done here could be used to attack it. Just not sure how. I figured out the coefficients of the polynomial in the numerator, and the partial decomposition one can use. $\endgroup$ Commented Aug 10, 2023 at 0:32
  • $\begingroup$ @user3108815 Haven't got time to investigate. $\endgroup$
    – oO_ƲRF_Oo
    Commented Aug 10, 2023 at 10:12

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