Prove that:$\int_{0}^{1}{1+x+x^2+\cdots+x^{4n-1}\over (1+x^{2n})(-\ln{x})^{1\over s}}dx=\Gamma\left({s-1\over s}\right)...$

Question

Show that

$$\int_{0}^{1}{1+x+x^2+x^3+\cdots+x^{4n-1}\over (1+x^{2n})(-\ln{x})^{1\over s}}dx=\Gamma\left({s-1\over s}\right)\sum_{k=1}^{2n}{1\over k^{s-1\over s}}\tag1$$

$s>1$ and $n\ge1$ [both are integers]

My try:

Let take $n=1$ and $s=3$ as an example

$$\int_{0}^{1}{1+x+x^2+x^3\over (1+x^2)(-\ln{x})^{1\over 3}}dx=\Gamma(2/3)\left(1+{1\over 2^{2\over 3}}\right)$$

$1+x+x^2+x^3=(1+x)(1+x^2)$

$$\int_{0}^{1}{1+x\over(-\ln{x})^{1\over 3}}dx$$

Can we apply the Frullani theorem at this point? OR making a sub:

$u=-\ln{x}$ then $e^{-u}=x$ and $du={-1\over x}dx$

$$\int_{0}^{\infty}{e^{-u}+e^{-2u}\over u^{1\over 3}}du$$

$$\int_{0}^{\infty}({e^{-u}u^{-1\over 3}+e^{-2u}u^{-1\over 3}})du$$

At this point we can apply integration by parts? Any way this is too long of a process of doing. Can anyone show me a general way of proving this integral (1)

Answer 1

One may first simplify the integrand observing that $$ {1+x+x^2+x^3+\cdots+x^{4n-1}\over (1+x^{2n})}=\frac{1}{1+x^{2n}}\cdot \frac{1-x^{4n}}{1-x}=\sum_{k=0}^{2n-1}x^k $$ then the given integral rewrites $$ \int_{0}^{1}{1+x+x^2+x^3+\cdots+x^{4n-1}\over (1+x^{2n})(-\ln{x})^{1\over s}}\:dx=\sum_{k=0}^{2n-1}\int_{0}^{1}\frac{x^k\:dx}{(-\ln{x})^{1/s}}=\Gamma\left(1-\frac1s\right)\sum_{k=1}^{2n}{1\over k^{1-1/s}} $$ as announced, since by the change of variable $u=-\ln{x}$, $x=e^{-u}$, one has $$ \int_{0}^{1}\frac{x^k\:dx}{(-\ln{x})^{1/s}}=\int_0^\infty u^{-1/s}e^{-(k+1)u}du=\Gamma\left(1-\frac1s\right)\cdot \frac1{(k+1)^{1-1/s}}. $$