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I wondered how to calculate an approximation of next integral involving the fractional part function $\{x\}=x-\lfloor x\rfloor$ and the so-called Gudermannian function denoted in this post as $\operatorname{gd} x$

$$\mathcal{G}=\int_0^{\log\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}}\{\frac{1}{\operatorname{gd} x}\}\,dx.\tag{1}$$

The upper limit is due that I think that is a simple upper limit for this integral, since if my calculations are rights

$$\mathcal{G}=\int_0^{\pi/4}\{\frac{1}{y}\}\frac{dy}{\cos y}=\int_{4/\pi}^{\infty}\frac{\{v\}}{v^2}\frac{dv}{\cos\frac{1}{v}}.\tag{2}$$

Question. Can you find an approximation for the positive real number $\mathcal{G}\,$? You can to use inequalities or series expansions and/or integration or a different method to calculate an approximation of $(1)$. Many thanks.

Using a CAS I know a very good approximation of previous definite integral, that I am asking about a method or reasoning to get an aproximation of our integral $\mathcal{G}$. I know simple inequalities that involve the cosine function (but you can use $(1)$ or $(2)$ in your approach).

For example, $\int_{4/\pi}^{\infty}\frac{1}{v^2}\frac{dv}{\cos\frac{1}{v}}$ is about twice times $\mathcal{G}$, thus isn't a good approximation for $\mathcal{G}$.

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  • $\begingroup$ I hope that my integral is well-received, my purpose is a curiosity, (in the context that I tried to get the closed-form of such integral $\mathcal{G}$ for a simple upper limit but seems impossible) about how get an approximation of our definite integral $\int_0^{\log\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}}\{\frac{1}{\operatorname{gd} x}\}\,dx$. Thus I am asking about an ingenious strategy to get an aproximation of the value for $\mathcal{G}$ that is provided for any CAS. Isn't required a very good approximation, just provide a remarkable calculations that provide us a good approximation. $\endgroup$
    – user243301
    Jun 6, 2018 at 9:09

1 Answer 1

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Using identity:

$$\lfloor x\rfloor =x+\frac{\tan ^{-1}(\cot (\pi x))}{\pi }-\frac{1}{2}$$

We can find:

$$\{\frac{1}{\operatorname{gd} x}\}=\frac{1}{2}-\frac{\tan ^{-1}\left(\cot \left(\frac{\pi }{\text{gd}(x)}\right)\right)}{\pi }$$

Expanding with series around $x=0$ with order=4

$$\{\frac{1}{\operatorname{gd} x}\}=\frac{1}{2}-\frac{\tan ^{-1}\left(\cot \left(\frac{\pi }{x}+\frac{\pi x}{6}-\frac{\pi x^3}{72}+\frac{43 \pi x^5}{15120}-\frac{271 \pi x^7}{362880}+O\left(x^9\right)\right)\right)}{\pi }$$

with CAS help like Mathematica 11.3:

 sol1 = NIntegrate[1/Gudermannian[x] - Floor[1/Gudermannian[x]], {x, 0, 
 Log[Sqrt[(2 + Sqrt[2])/(2 - Sqrt[2])]]}, Method -> "LocalAdaptive"];

 sol2 = NIntegrate[1/2 - ArcTan[Cot[\[Pi]/x + (\[Pi] x)/6 - (\[Pi] x^3)/72 + (43 \[Pi] x^5)/
 15120 - (271 \[Pi] x^7)/362880]]/\[Pi], {x, 0, Log[Sqrt[(2 + Sqrt[2])/(2 - Sqrt[2])]]}, 
 Method -> "LocalAdaptive"];
 Abs[(sol1 - sol2)/sol1]

absolue error is about: $0.0000110693$.

EDITED:

If we only take 1 term then:

 Integrate[1/2 - ArcTan[Cot[Pi/x]]/Pi, {x, 0, Log[Sqrt[(2 + Sqrt[2])/(2 - Sqrt[2])]]}]
 (*Gives: Infinity ->a  Bug !!!*)

Numerics gives:

 NIntegrate[1/2 - ArcTan[Cot[Pi/x]]/Pi, {x, 0, 
 Log[Sqrt[(2 + Sqrt[2])/(2 - Sqrt[2])]]}, Method -> "LocalAdaptive",PrecisionGoal -> 25]
 (* 0.415137 *)

Calculating in Maple 2018. Changing to x=1/t

 int((Pi-2*arctan(cot(Pi*t)))/(2*Pi*t^2), t = 2/ln(3+2*sqrt(2)) .. infinity);

$$\int_0^{\ln \left(\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}\right)} \left(\frac{1}{2}-\frac{\tan ^{-1}\left(\cot \left(\frac{\pi }{x}\right)\right)}{\pi }\right) \, dx=\int_{2\, \left( \ln \left( 3+2\,\sqrt {2} \right) \right) ^{-1}}^{ \infty }\!{\frac {{\rm arccot} \left(\cot \left( \pi\,t \right) \right)}{\pi\,{t}^{2}}}\,{\rm d}t =2-\gamma -\frac{1}{2} \ln \left(3+2 \sqrt{2}\right)+\ln \left(\frac{1}{2} \ln \left(3+2 \sqrt{2}\right)\right)\approx 0.415137$$

where $\gamma$ is Euler–Mascheroni constant

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  • $\begingroup$ Many thanks for your answer, I am going to study the details. $\endgroup$
    – user243301
    Jun 6, 2018 at 9:48

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