3
$\begingroup$

It is well-known

$$\int_{0}^{1}\left({1\over \ln(x)}+{1\over 1-x}\right)\mathrm dx=\gamma\tag1$$

Messing around with $(1)$ using wolfram integrator, we have

$$\int_{0}^{1}\left({x^u\over \ln(x)}+{x^v\over 1-x}\right)\mathrm dx=F(u,v)\tag2$$

$H_0=0.$

How do we show that $F(u,v)=\gamma-H_v+\ln(u+1)?$

$\endgroup$
4
$\begingroup$

Use \begin{eqnarray*} H_v= \int_0^1 \frac{1-x^v}{1-x} dx \end{eqnarray*} and \begin{eqnarray*} \frac{x^u-1}{\ln x} = \int_0^u x^t dt. \end{eqnarray*} We have \begin{eqnarray*} \int_0^1 ( \frac{x^u}{\ln x}+ \frac{x^v}{1-x}) dx &=& \int_0^1 ( \frac{1}{\ln x}+ \frac{1}{1-x}- \frac{1-x^v}{1-x}+\frac{x^u-1}{\ln x}) dx \end{eqnarray*} The first two terms of the RHS give $ \gamma $ , the third term gives $H_v$ and the fourth term is \begin{eqnarray*} \int_0^1 \frac{x^u-1}{\ln x} dx = \int_0^1 \int_0^u x^t dt dx = \int_0^u \int_0^1 x^t dx dt = \int_0^u \frac{1}{1+t} dt = \ln (1+u). \end{eqnarray*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.