It is know that $$H_n=-n\int_0^1(1-t)^{n-1}\log (t)dt,$$ see [1], where $H_n=1+1/2+\ldots+1/n$ it the nth harmonic number. Then I believe that can be used for $x>0$ $$\frac{1}{2}\log(x)=\sum_{k=1}^{\infty}\frac{(-1)^{2k-1}}{2k-1}\left(\frac{x-1}{1+x}\right)^{2k-1},$$ see for example [2], to show $$H_n=2n\sum_{k=1}^{\infty}\frac{1}{2k-1}\int_0^1\frac{(1-x)^{n+2k-2}}{(1+x)^{2k-1}}dx.$$

Question. It is possible to get a closed-form for $$\int_0^1\frac{(1-x)^{n+2k-2}}{(1+x)^{2k-1}}dx?$$

If you want give a justification for previous expression for $H_n$, and know how comoute previous definite integral I am wait your answer. Thanks in advance.


[1] Furdui, LA GACETA de la Real Sociedad Matemática Española. Third paragraph in page 699, see here.

[2] Hyslop, Infinite Series, Dover Publications (2006).

  • $\begingroup$ My attempt was to use Wolfram Alpha. I can not evaluate this integral with its definite integral calculator. A evaluation of the integral is provide us in terms of a hypergeometric function. $\endgroup$ – user243301 Jan 19 '16 at 22:52

(More of a comment, since I think you suggest you know this, but maybe this is useful to somebody else answering.)

According to Mathematica if $k,n\in\mathbb{Z}$ and $2 k+n>1$,

$$I=\frac{\, _2F_1(1,2 k-1;2 k+n;-1)}{2 k+n-1}$$

Where $F$ is the hypergeometric function on MathWorld.

  • $\begingroup$ Very thanks much for your answer @GBeau I didn't known this, I've used another integrator and the expression given as output was more difficult to evaluate. $\endgroup$ – user243301 Jan 20 '16 at 8:01

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