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Using the closed-form that provide us Wolfram Alpha of the similar definite integral involving the Euler-Mascheroni constant $$\int_0^\infty e^{-x}\log^3x dx $$

integrate e^(-x)log^3(x)dx, from x=0 to infinite

one can calculate when do the change of variable $x=e^u$ that $$\int_0^\infty \sum_{k=0}^\infty \frac{(-1)^k}{k!}e^{(k+1)u}u^3 du-6-6\sum_{k=1}^\infty\frac{(-1)^k}{k!(k+1)^4}=-2\zeta(3)-\gamma^3-\frac{\gamma\pi^2}{2}.$$

After calculate by integration by parts the integral in LHS, and one more time using Wolfram Alpha one can write that

$$3\int_0^\infty x^2 e^{-e^x}dx=3\cdot 2\cdot G_{3,4}^{4,0}\left(1\left|\begin{smallmatrix}1,1,1\\ 0,0,0,0\end{smallmatrix}\right.\right),$$ where this RHS is a particular value of the Meijer G-Function, you can see with this code

integrate x^2 e^(-e^x)dx, from x=0 to infinite.

And finally I've calculated with Wolfram Alfa

$$\sum_{k=1}^\infty \frac{(-1)^k}{k! (k+1)^4}=_4F_4(1,1,1,1;2,2,2,2;-1)-1,$$

where in RHS is a particular value of a hypergeometric, see if you need it

sum (-1)^k/((k+1)^4 k!), from k=1 to infinite.

I would like to clarify if were justified all steps, is not neccesary all calculations, only if were justified, and if there are mistakes, because I've doubts.

Question. Can you provide us a summary as a guideline about how justify the more important steps, those in which could have more doubts? Aren't required the calculations in detail. Many thanks.

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  • $\begingroup$ I understand that using these special functions I can not get a closed form, but if you know something about previous special functions to get a more interesting identity add a comment or addendum. Thanks for the attention and patience! $\endgroup$ – user243301 Dec 7 '16 at 3:08
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Do you want to know something about Apéry's constant and Euler-Mascheroni constant, Meijer G-Function and a hypergeometric or simply solve the integral ?

For the first case I cannot help you well (sorry) but for the second case I continue as follows:

Be $ \enspace\displaystyle F(z):=\Gamma(1+z)=\lim\limits_{n\to\infty}\frac{n^z}{\prod\limits_{k=1}^n\left(1+\frac{z}{k}\right)}$ . $\hspace{0.5cm}$Value range: It's enough to choose real $z>-1$ .

It’s $\enspace \displaystyle \frac{d}{dz}\ln F(z)|_{z=0}=-\gamma $

and $\enspace\displaystyle \frac{d^k}{dz^k}\ln F(z)|_{z=0}=(-1)^k(k-1)!\zeta(k)\enspace$ for natural number $\enspace k\geq 2$ .

And it’s $\enspace \displaystyle \sum\limits_{k=0}^\infty \frac{z^k}{k!} \int\limits_0^\infty e^{-x}(\ln x)^k dx= \int\limits_0^\infty e^{-x} \sum\limits_{k=0}^\infty \frac{(z\ln x)^k}{k!}dx= \int\limits_0^\infty e^{-x}x^z dx=F(z) $

which leads to $\enspace\displaystyle \int\limits_0^\infty e^{-x}(\ln x)^k dx= \frac{d^k}{dz^k} F(z)|_{z=0}$ . $\hspace{2cm}$ (Taylorseries)

For the case $\enspace k=3\enspace $ we get

$\displaystyle \int\limits_0^\infty e^{-x}(\ln x)^3 dx= F’’’(z)|_{z=0}=$

$\displaystyle =F(z)( (\ln F(z))’^3 +3(\ln F(z))’ (\ln F(z))’’+2(\ln F(z))’’’ )|_{z=0}=-\gamma^3-3\gamma\zeta(2)-2\zeta(3)$ .

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  • $\begingroup$ Many thanks for your answer, this afternoon I am going to understand your nice calculations. $\endgroup$ – user243301 Dec 7 '16 at 13:22
  • $\begingroup$ @user243301 : You are welcome. :-) $\endgroup$ – user90369 Dec 7 '16 at 13:24
  • $\begingroup$ Which are welcome are users as you with nice answers in his/her hands, as this. Thanks!!! $\endgroup$ – user243301 Dec 7 '16 at 13:28
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    $\begingroup$ @user243301 In other words you need $\frac{\Gamma'(s)}{\Gamma(s)} = -\gamma+\sum_{n=0}^\infty \frac{1}{n+1}-\frac{1}{s+n}$ whose $k$th derivative at $s =1$ is something $k! \zeta(k+1)$ $\endgroup$ – reuns Dec 7 '16 at 17:05
  • $\begingroup$ You are welcome @user1952009, many thanks for your claims, now I try think about it. Merci. $\endgroup$ – user243301 Dec 7 '16 at 17:07

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