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Find: $$I=\int_0^{1}\frac{(1+x^2)\ln(1+x^4)}{1+x^4}\,dx$$

I need to use the definition of do Trigamma function but I don't see how.

My try was as following :

Let $y=x^4$ then $dy=4x^{3}\,dx$ so $\;dx=\frac{dy}{4y^{\frac{3}{4}}}$.

$$I=\int_0^{1}\frac{(1+x^{\frac{3}{4}})\ln(1+x)}{x^{\frac{3}{4}}+x^{\frac{7}{4}}}dx$$

Use : $\ln(1+x)=-\sum_{n=1}^{\infty}\frac{(-1)^{n}x^n}{n}$

We obtain

$$J=\int_0^{1}\frac{x^n+x^{(3+4n)/4}}{x^{3/4}+x^{7/4}}dx$$

Then I use definition of digamma function

$$\int_0^{1}\frac{x^{a-1}\ln x}{1+x}\,dx=\beta'(a)$$

Where: $$2\beta(a)=\psi\left(\frac{1+a}{2}\right)-\psi\left(\frac{a}{2}\right)$$

But I find sum related with digamma function.

So I don't have other ideas to approach it!

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  • $\begingroup$ Which definition of the Trigamma Function? $\endgroup$ – mrtaurho Jul 1 at 11:42
  • $\begingroup$ See now I add soothing !! $\endgroup$ – Kînan Jœd Jul 1 at 11:51
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    $\begingroup$ The result of MMA is very large! $\endgroup$ – Dr. Sonnhard Graubner Jul 1 at 11:52
  • $\begingroup$ @Dr.SonnhardGraubner. The result of MMA is very small !. MMA code: -Derivative[0, 1, 0, 0][Hypergeometric2F1][1/4, 1, 5/4, -1] - Derivative[0, 1, 0, 0][Hypergeometric2F1][3/4, 1, 7/4, -1]/3 $\endgroup$ – Mariusz Iwaniuk Jul 1 at 12:31
  • $\begingroup$ Ok, i'm using an older version $\endgroup$ – Dr. Sonnhard Graubner Jul 1 at 12:33
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$$\small \boxed{\int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx=\frac{3\pi}{2\sqrt 2}\ln 2-\frac1{32}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right)}$$ To see that write the integral as: $$I=\int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx\overset{x\to \frac{1}{x}}=\int_1^\infty \frac{(1+x^2) (\ln(1+x^4)-\ln(x^4))}{1+x^4}dx$$ $$\Rightarrow 2I=\int_0^\infty \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx-4\int_1^\infty \frac{(1+x^2)\ln x}{1+x^4}dx$$


For the first integral consider: $$I(a)=\int_0^\infty \frac{(1+x^2)\ln(a+x^4)}{1+x^4}dx\Rightarrow I'(a)=\int_0^\infty \frac{1+x^2}{(a+x^4)(1+x^4)}dx$$ $$=\frac{1}{a-1}\int_0^\infty \frac{1+x^2}{1+x^4}dx-\frac{1}{a-1}\int_0^\infty \frac{1+x^2}{a+x^4}dx$$ $$\int_0^\infty \frac{1+x^2}{1+x^4}dx=\int_0^\infty \frac{1+\frac{1}{x^2}}{\left(x-\frac{1}{x}\right)^2+2}=\frac{1}{\sqrt 2}\arctan\left(\frac{x^2-1}{\sqrt 2 x}\right)\bigg|_0^\infty =\frac{\pi}{\sqrt 2}$$ $$\int_0^\infty \frac{1+x^2}{a+x^4}dx\overset{x=\sqrt[4]at}=\sqrt[4]a\int_0^\infty \frac{1+\sqrt at^2}{a(1+t^4)}dt=a^{-3/4}\int_0^\infty \frac{dt}{1+t^4}+a^{-1/4}\int_0^\infty \frac{t^2dt}{1+t^4}$$ $$\int_0^\infty \frac{1}{1+t^4}dt\overset{t\to \frac{1}{t}}=\int_0^\infty \frac{t^2}{1+t^4}dt=\frac12 \int_0^\infty \frac{1+t^2}{1+t^4}dt=\frac{\pi}{2\sqrt 2}$$ $$\Rightarrow I'(a)= \frac{\pi}{2\sqrt{2}}\left(\frac{2-a^{-1/4}-a^{-3/4}}{a-1}\right) $$ We are looking here for $I(1)$, but $I(0)=0$, (put $x=\frac{1}{x}$ to see that), so: $$\Rightarrow I(1)= \frac{\pi}{2\sqrt{2}}\int_0^1 \left(\frac{1-a^{-1/4}+1-a^{-3/4}}{a-1}\right)da=\frac{3\pi}{\sqrt 2}\ln 2 $$ Above follows by splitting into two parts and using this.


For the second integral we will use the trigamma function as you mentioned in your question. $$\int_1^\infty \frac{(1+x^2)\ln x}{1+x^4}dx\overset{x\to \frac{1}{x}}=-\int_0^1 \frac{\ln x}{1+x^4}dx-\int_0^1 \frac{x^2\ln x}{1+x^4}dx$$ $$=\frac1{64} \left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right)$$ One can use trigamma's reflection formula in order to simplify two more terms, but I think it looks nice this way.

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    $\begingroup$ (+1) Feynman; again xD $\endgroup$ – mrtaurho Jul 1 at 13:36
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(Too long for a comment.)

Zacky gave the answer:

$$\small \int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx=\frac{3\pi}{2\sqrt 2}\ln 2-\frac1{32}\left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right)$$

After my post on Catalan's constant $K = \rm{Cl}_2\big(\tfrac\pi2\big)$ and Gieseking's constant $\kappa =\rm{Cl}_2\big(\tfrac\pi3\big)$, I recognized the four polygammas as involving Catalan's quartic counterpart $\rm{Cl}_2\big(\color{red}{\tfrac\pi4}\big)$, namely,

$$-\small\frac1{32} \left(\psi_1\left(\frac{1}{8}\right)+\psi_1\left(\frac{3}{8}\right)-\psi_1\left(\frac{5}{8}\right)-\psi_1\left(\frac{7}{8}\right)\right) = \frac1{\sqrt2}\Big(K-4\,\rm{Cl}_2\big(\color{red}{\tfrac\pi4}\big)\Big)$$

Hence we have the simple closed-form,

$$\int_0^1 \frac{(1+x^2)\ln(1+x^4)}{1+x^4}dx=\frac{3\pi}{2\sqrt2}\ln 2+ \frac K{\sqrt2} -\frac{4\rm{Cl}_2\big(\color{red}{\tfrac\pi4}\big)}{\sqrt2}$$

where $\rm{Cl}_2\big(\tfrac\pi4\big) = \Im\, \rm{Li}_2(e^{2\pi i/8})$.

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