Evaluate the integral:
$$\mathcal{J} =\int_0^1 \left( \frac{1}{1-x} + \frac{1}{\ln x} \right) \, {\rm d}x$$
Solution
We begin with the very well known identity:
$$\int_{0}^{1} \frac{1-t^x}{1-t} \, {\rm d}t = \mathcal{H}_x$$
Now , let's see what shall happen if we actually integrate it. Successively we have
\begin{align*}
\int_{0}^{1}\mathcal{H}_x \, {\rm d}x &= \int_{0}^{1} \int_{0}^{1} \frac{1-t^x}{1-t} \,{\rm d}t \, {\rm d}x \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1-t^x}{1-t} \, {\rm d}x \, {\rm d}t\\
&= \int_{0}^{1}\frac{1}{1-t} \int_{0}^{1}\left [ 1-t^x \right ] \, {\rm d}x \, {\rm d}t\\
&= \int_{0}^{1}\left (\frac{1}{1-t} \left [ x- \frac{t^x}{ \ln t} \right ]_0^1 \right ) \, {\rm d}t \\
&= \int_{0}^{1} \left ( \frac{1}{1-t} + \frac{1}{\ln t} \right ) \, {\rm d}t
\end{align*}
Aha! Magic. Our integral appeared. Hmm.. the main problem actually is how to evaluate the integral of the harmonic number. Hm! That's not so difficult. Recall the following:
\begin{equation}\mathcal{H}_x = \gamma + \psi(x+1) \end{equation}
where $\psi$ stands for the digamma function. Thus , integrating $(1)$ we have that:
\begin{align*}
\int_{0}^{1} \mathcal{H}_x \, {\rm d}x &=\int_{0}^{1} \left [ \gamma + \psi(x+1) \right ] \, {\rm d}x \\
&= \left [ \gamma x + \log \Gamma(x+1) \right ]_0^1\\
&= \gamma + \log \Gamma(2) - \log \Gamma(1)\\
&= \gamma
\end{align*}
where $\gamma$ stands for the Euler - Mascheroni constant. In fact, it holds that:
\begin{equation}\int_0^1 \left ( \frac{1}{1-x} + \frac{1}{\ln x} \right ) \, {\rm d}x = \lim_{s\rightarrow 1} \left ( \zeta(s) - \frac{1}{s-1} \right ) \end{equation}
$$\mathcal{J} =\int_0^1 \left( \frac{1}{1-x} + \frac{1}{\ln x} \right) \, {\rm d}x$$
Solution
We begin with the very well known identity:
$$\int_{0}^{1} \frac{1-t^x}{1-t} \, {\rm d}t = \mathcal{H}_x$$
Now , let's see what shall happen if we actually integrate it. Successively we have
\begin{align*}
\int_{0}^{1}\mathcal{H}_x \, {\rm d}x &= \int_{0}^{1} \int_{0}^{1} \frac{1-t^x}{1-t} \,{\rm d}t \, {\rm d}x \\
&= \int_{0}^{1} \int_{0}^{1} \frac{1-t^x}{1-t} \, {\rm d}x \, {\rm d}t\\
&= \int_{0}^{1}\frac{1}{1-t} \int_{0}^{1}\left [ 1-t^x \right ] \, {\rm d}x \, {\rm d}t\\
&= \int_{0}^{1}\left (\frac{1}{1-t} \left [ x- \frac{t^x}{ \ln t} \right ]_0^1 \right ) \, {\rm d}t \\
&= \int_{0}^{1} \left ( \frac{1}{1-t} + \frac{1}{\ln t} \right ) \, {\rm d}t
\end{align*}
Aha! Magic. Our integral appeared. Hmm.. the main problem actually is how to evaluate the integral of the harmonic number. Hm! That's not so difficult. Recall the following:
\begin{equation}\mathcal{H}_x = \gamma + \psi(x+1) \end{equation}
where $\psi$ stands for the digamma function. Thus , integrating $(1)$ we have that:
\begin{align*}
\int_{0}^{1} \mathcal{H}_x \, {\rm d}x &=\int_{0}^{1} \left [ \gamma + \psi(x+1) \right ] \, {\rm d}x \\
&= \left [ \gamma x + \log \Gamma(x+1) \right ]_0^1\\
&= \gamma + \log \Gamma(2) - \log \Gamma(1)\\
&= \gamma
\end{align*}
where $\gamma$ stands for the Euler - Mascheroni constant. In fact, it holds that:
\begin{equation}\int_0^1 \left ( \frac{1}{1-x} + \frac{1}{\ln x} \right ) \, {\rm d}x = \lim_{s\rightarrow 1} \left ( \zeta(s) - \frac{1}{s-1} \right ) \end{equation}
Shall we continue a bit? The following hold:
ReplyDelete1.$\displaystyle \int_{0}^{1} \left( \frac{1}{\log x} + \frac{1}{1-x} \right)^2 \, {\rm d}x = \log (2\pi) - \frac{3}{2}$. See here for a proof.
2. $\displaystyle \int_{0}^{1}\left ( \frac{1}{\log x} + \frac{1}{1-x} \right )^3 \, {\rm d}x = - \frac{31}{24} + 6 \log \mathbb{A}$ where $\mathbb{A}$ stands for the Glashier - Kinkelin constant.
3. Now comes the hilarious part:
In general for $m \geq 2$ it holds that:
\begin{align*}
& \int_{0}^{1}\left( \frac{1}{\log x} + \frac{1}{1-x} \right)^{m} \, dx \\
&= -\frac{H_{m-1}}{(m-1)!} + \frac{1}{(m-1)!} \sum_{k=1}^{m-1} \left[{{m-1}\atop{k}}\right] \zeta(1-k) \\
&\quad - \frac{1}{(m-2)!}\sum_{j=1}^{m-1}\sum_{l=m-j}^{m-1} \binom{m}{j} \binom{m-2}{j-1} \left[{{j-1}\atop{l+j-m}}\right] \{ \zeta'(1-l) + H_{m-j-1} \zeta(1-l) \}
\end{align*}
Shock! I , personally, have no idea how to prove this.
An extension of the original integral is the following:
\begin{align*}
\gamma &= \int_{0}^{1}\left ( \frac{1}{1-x} + \frac{1}{\ln x} \right ) \, {\rm d}x \\
&\!\!\!\!\overset{x=e^{-y}}{=\! =\! =\! =\!} \int_{0}^{\infty} \left ( \frac{e^{-y}}{1-e^{-y}} - \frac{e^{-y}}{y} \right ) \, {\rm d}y
\end{align*}
Of course we already knew about that since in the digamma wiki page the formula is already stated under the Integral Represantations tab.