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show that:

$$1+q\int_{0}^{1}x^{1-qx}dx=\sum_{k=0}^{\infty}\left(\dfrac{q}{k+1}\right)^k\cdots\cdots(1)$$

I kown prove following this

$$\int_{0}^{1}x^{-qx}dx=\sum_{n=0}^{\infty}\dfrac{q^n}{(n+1)^{n+1}}$$

note that $$\int_{0}^{1}x^{-qx}=\int_{0}^{1}e^{qx\ln{\frac{1}{x}}}dx=\sum_{n=0}^{\infty}\frac{q^n}{n!}\int_{0}^{1}x^n\ln^n{\frac{1}{x}}dx$$

let $$\ln{\dfrac{1}{x}}=t\Longrightarrow \int_{0}^{1}x^n\ln^n{\dfrac{1}{x}}dx=\int_{0}^{\infty}e^{-(n+1)t}t^ndt=\dfrac{\Gamma{(n+1)}}{(n+1)^{n+1}}$$ so

$$\int_{0}^{1}x^{-qx}dx=\sum_{n=0}^{\infty}\dfrac{q^n}{(n+1)^{n+1}}$$

But this $(1)$ I can't prove it,Thank you everyone.

Cameron Williams
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math110
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  • Hint: the Maple command $$series(x^{-q*x+1}, x, 6) $$ produces $$ x-q\ln \left( x \right) {x}^{2}+1/2,{q}^{2} \left( \ln \left( x \right) \right) ^{2}{x}^{3}-1/6,{q}^{3} \left( \ln \left( x \right) \right) ^{3}{x}^{4}+1/24,{q}^{4} \left( \ln \left( x \right) \right) ^{4}{x}^{5}+O \left( {x}^{6} \right) $$ – user64494 Sep 16 '13 at 17:28

1 Answers1

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The proof goes along the same lines $$ \begin{align} \int_0^1 x^{1-qx}dx &=\int_0^1 xe^{-qx}dx\\ &=\int_0^1 x\sum\limits_{n=0}^\infty \frac{(qx\ln x^{-1})^n}{n!}dx\\ &=\sum\limits_{n=0}^\infty\frac{q^n}{n!}\int_0^1 x^{n+1}\ln^n x^{-1}dx\qquad\{\ln x^{-1}=t\}\\ &=\sum\limits_{n=0}^\infty\frac{q^n}{n!}\int_{+\infty}^0 e^{-(n+1)t}t^n (-e^{-t})dt\\ &=\sum\limits_{n=0}^\infty\frac{q^n}{n!}\int_0^\infty e^{-(n+2)t}t^n dt\qquad \{(n+2)t=s\}\\ &=\sum\limits_{n=0}^\infty\frac{q^n}{n!}\int_0^\infty e^{-s}(n+2)^{-n}s^n (n+2)^{-1}ds\\ &=\sum\limits_{n=0}^\infty\frac{q^n}{(n+2)^{n+1}} \end{align} $$ Thus $$ \begin{align} 1+q\int_0^1 x^{1-qx}dx &=1+q\sum\limits_{n=0}^\infty\frac{q^n}{(n+2)^{n+1}}\\ &=1+\sum\limits_{n=0}^\infty\frac{q^{n+1}}{(n+2)^{n+1}}\qquad\{k=n+1\}\\ &=\frac{q^0}{(0+1)^0}+\sum\limits_{k=1}^\infty\frac{q^k}{(k+1)^k}\\ &=\sum\limits_{k=0}^\infty\left(\frac{q}{k+1}\right)^k\\ \end{align} $$

Norbert
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  • Aw man. I just started typing up my answer (which I had just finished before class an hour ago but didn't have time to type up). Good job though! – Cameron Williams Sep 16 '13 at 18:31
  • My apologies,for being the first. Such things often happened to me, but not this time :-) – Norbert Sep 16 '13 at 18:34