Prove that $9\int_{0}^{\infty}x^5e^{-x^3}\ln(1+x)dx=\Gamma\left({1\over 3}\right)-\Gamma\left({2\over 3}\right)+\Gamma\left({3\over 3}\right)?$

Question

My last observance of this question by @Brightsun

Not so interesting integral but does have a neat closed form

$$9\int_{0}^{\infty}x^5e^{-x^3}\ln(1+x)\mathrm dx=\Gamma\left({1\over 3}\right)-\Gamma\left({2\over 3}\right)+\Gamma\left({3\over 3}\right)\tag1$$

An attempt:

$u=1+x$, then $(1)$ becomes

$$\int_{1}^{\infty}(u-1)^5e^{-(u-1)^3}\ln u\mathrm du\tag2$$

With this function $e^{-x^3}$ in $(1)$, it makes very hard to appl integratin by parts.

Recall $$\Gamma(t)=\int_{0}^{\infty}x^{t-1}e^{-x}\mathrm dx\tag3$$

Differentiate w.r.t t

$$\Gamma^{'}(t)=\int_{0}^{\infty}x^{t-1}\color{red}{e^{-x}}\ln x\mathrm dx\tag4$$ The problem is that the red part is not $e^{-x^3}$

Another attempt:

Recall $${1\over k}\Gamma(t)=\int_{0}^{\infty}x^{kt-1}e^{-x^k}\mathrm dx\tag5$$

Differentiate $(5)$ w.r.t t

$${1\over k^2}\Gamma^{'}(t)=\int_{0}^{\infty}x^{kt-1}e^{-x^k}\ln x\mathrm dx\tag6$$

Setting $k=3$ and $t=2$ we have

$$9\int_{0}^{\infty}x^5e^{-x^3}\ln(x)=\Gamma^{'}(2)\tag7$$

Not sure what is $\Gamma^{'}(2)?$

How else can we tackle $(1)?$

Answer 1

With a substitution the evaluation of $(1)$ boils down to the evaluation of $$ 3\int_{0}^{+\infty} x e^{-x} \log(1+x^{1/3})\,dx \stackrel{IBP}{=} \int_{0}^{+\infty}\frac{x+1}{x^{2/3}+x}e^{-x}\,dx\tag{A}$$ and with the inverse substitution the RHS of $(A)$ turns into $$ 3\int_{0}^{+\infty}(x^2-x+1)\,e^{-x^3}\,dx = \Gamma\left(\frac{3}{3}\right)-\Gamma\left(\frac{2}{3}\right)+\Gamma\left(\frac{1}{3}\right)\tag{B}$$ as wanted, pretty simple. As an alternative, one may notice that $$ \frac{x+1}{x^{2/3}+x}=\frac{(x^{1/3}+1)(x^{2/3}-x^{1/3}+1)}{(x^{1/3}+1)x^{2/3}}=1-x^{-1/3}+x^{-2/3}\tag{C}$$ and conclude from $(A)$ directly. Thanks to mickep.

Answer 2

As a generalization for Jack's method for $n$ is odd

$$I(n) = \int^\infty_0\frac{1+x^{-1}}{1+x^{-\frac{1}{n}}}e^{-x}\,dx =\sum_{k=0}^{n-1}(-1)^k \int^\infty_0 x^{-k/n}e^{-x}\,dx = \sum_{k=0}^{n-1}(-1)^k \Gamma\left(1-\frac{k}{n}\right) $$

One can rewrite using the reflection formula $$I(n)=\int^\infty_0\frac{1+x^{-1}}{1+x^{-\frac{1}{n}}}e^{-x}\,dx=1+\pi\sum_{k=1}^{n-1} \frac{(-1)^k \csc\left(k\pi/n \right)}{\Gamma(k/n)}$$