Evaluate the integral:
$$\mathcal{J}=\int_0^1 \frac{1-x}{(x+1) \log x}\, {\rm d}x$$
Solution
Let us consider the function $\displaystyle f(a)=\int_0^1 \frac{1-x^a}{(x+1) \ln x} \, {\rm d}x, \; a \geq 0$. Differentiting with respect to $a$ we have that:
\begin{align*}
f'(a) &=\frac{\mathrm{d} }{\mathrm{d} a} \int_{0}^{1} \frac{1-x^a}{(x+1) \ln x} \, {\rm d}x \\
&= \int_{0}^{1}\frac{\partial }{\partial a} \frac{1-x^a}{(x+1) \ln x} \, {\rm d}x\\
&= -\int_{0}^{1} \frac{x^a \log x}{(x+1) \log x} \, {\rm d}x\\
&= -\left [ x^a \ln (x+1) \right ]_0^1 +a \int_{0}^{1} x^{a-1} \ln (x+1) \, {\rm d}x \\
&=- \ln 2 + a \int_{0}^{1} x^{a-1} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} \, {\rm d}x \\
&= - \ln 2 +a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^{n+a-1} \, {\rm d}x \\
&= - \ln 2 + a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+a)} \\
&= -\ln 2 + a \left [ \frac{\psi^{(0)} \left ( \frac{a+1}{2} \right ) - \psi^{(0)} \left ( \frac{a+2}{2} \right )+2 \ln 2}{2a} \right ] \\
&=-\ln 2 + \frac{\psi^{(0)} \left ( \frac{a+1}{2} \right ) - \psi^{(0)} \left ( \frac{a+2}{2} \right )+2 \ln 2}{2} \\
&=\frac{\psi^{(0)} \left ( \frac{a+1}{2} \right ) - \psi^{(0)} \left ( \frac{a+2}{2} \right )}{2}
\end{align*}
Integrating back we have that
$$f(a)= \log \Gamma \left ( \frac{a+1}{2} \right ) -\log \Gamma \left ( \frac{a+2}{2} \right )+ c $$
Since $f(0)=0$ we get that $c=-\frac{\log \pi}{2}$. Thus:
\begin{align*}
f(1) &= -\log \Gamma \left ( \frac{3}{2} \right ) - \frac{\log \pi}{2}\\
&= - \log \frac{\sqrt{\pi}}{2} - \frac{\log \pi}{2}\\
&= -\log \sqrt{\pi} + \log 2 - \frac{\log \pi}{2} \\
&= - \frac{1}{2} \log \pi + \log 2 - \frac{\log \pi}{2} \\
&= \log 2 - \log \pi \\
&= \log \frac{2}{\pi}
\end{align*}
Therefore we conclude that:
$$\int_{0}^{1}\frac{1-x}{(x+1) \log x} \, {\rm d}x = \log \left ( \frac{2}{\pi} \right )$$
$$\mathcal{J}=\int_0^1 \frac{1-x}{(x+1) \log x}\, {\rm d}x$$
Solution
Let us consider the function $\displaystyle f(a)=\int_0^1 \frac{1-x^a}{(x+1) \ln x} \, {\rm d}x, \; a \geq 0$. Differentiting with respect to $a$ we have that:
\begin{align*}
f'(a) &=\frac{\mathrm{d} }{\mathrm{d} a} \int_{0}^{1} \frac{1-x^a}{(x+1) \ln x} \, {\rm d}x \\
&= \int_{0}^{1}\frac{\partial }{\partial a} \frac{1-x^a}{(x+1) \ln x} \, {\rm d}x\\
&= -\int_{0}^{1} \frac{x^a \log x}{(x+1) \log x} \, {\rm d}x\\
&= -\left [ x^a \ln (x+1) \right ]_0^1 +a \int_{0}^{1} x^{a-1} \ln (x+1) \, {\rm d}x \\
&=- \ln 2 + a \int_{0}^{1} x^{a-1} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^n}{n} \, {\rm d}x \\
&= - \ln 2 +a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1} x^{n+a-1} \, {\rm d}x \\
&= - \ln 2 + a \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n(n+a)} \\
&= -\ln 2 + a \left [ \frac{\psi^{(0)} \left ( \frac{a+1}{2} \right ) - \psi^{(0)} \left ( \frac{a+2}{2} \right )+2 \ln 2}{2a} \right ] \\
&=-\ln 2 + \frac{\psi^{(0)} \left ( \frac{a+1}{2} \right ) - \psi^{(0)} \left ( \frac{a+2}{2} \right )+2 \ln 2}{2} \\
&=\frac{\psi^{(0)} \left ( \frac{a+1}{2} \right ) - \psi^{(0)} \left ( \frac{a+2}{2} \right )}{2}
\end{align*}
Integrating back we have that
$$f(a)= \log \Gamma \left ( \frac{a+1}{2} \right ) -\log \Gamma \left ( \frac{a+2}{2} \right )+ c $$
Since $f(0)=0$ we get that $c=-\frac{\log \pi}{2}$. Thus:
\begin{align*}
f(1) &= -\log \Gamma \left ( \frac{3}{2} \right ) - \frac{\log \pi}{2}\\
&= - \log \frac{\sqrt{\pi}}{2} - \frac{\log \pi}{2}\\
&= -\log \sqrt{\pi} + \log 2 - \frac{\log \pi}{2} \\
&= - \frac{1}{2} \log \pi + \log 2 - \frac{\log \pi}{2} \\
&= \log 2 - \log \pi \\
&= \log \frac{2}{\pi}
\end{align*}
Therefore we conclude that:
$$\int_{0}^{1}\frac{1-x}{(x+1) \log x} \, {\rm d}x = \log \left ( \frac{2}{\pi} \right )$$
Here is a very similiar problem but it is much more challenging.
ReplyDeleteProblem: Evaluate the integral:
$$\int_0^1 \frac{x\ln x}{\ln (1-x)} \, {\rm d}x$$
Solution
Start things off by applying the sub $u=1-x$. Thus the integral is transformed to:
$$\int_{0}^{1} \frac{x \ln x}{\ln (1-x)} \, {\rm d}x \overset{u=1-x}{=\! =\! =\! =\! } \int_{0}^{1} \frac{(1-u) \ln (1-u)}{\ln u} \, {\rm d}u$$
Now consider the function $f(a)=\int_0^1 \frac{(1-u^a) \ln (1-u)}{\ln u} \, {\rm d}u$. Cleary $f(0)=0$. Now differentiate with respect to $a$ once, thus:
\begin{align*}
f'(a) &= \frac{\mathrm{d} }{\mathrm{d} a} \int_{0}^{1} \frac{(1-u^a) \ln (1-u)}{\ln u} \, {\rm d}u\\
&= \int_{0}^{1} \frac{\partial }{\partial a} \frac{(1-u^a) \ln (1-u)}{\ln u} \, {\rm d}u\\
&=- \int_{0}^{1} u^a \ln (1-u) \, {\rm d}u \\
&= \frac{\mathcal{H}_{a+1}}{a+1}\\
&= \frac{1}{(a+1)} \left ( \psi^{(0)} (a+2) + \gamma \right )
\end{align*}
Now we have to evaluate $f(1)$. For that we have that:
\begin{align*}
f(1) &= \int_{0}^{1} f'(a) \, {\rm d}a \\
&= \int_{0}^{1}\frac{1}{a+1} \left ( \psi^{(0)} (a+2) + \gamma \right ) \, {\rm d}a\\
&= \int_{0}^{1} \frac{1}{a+1} \left ( -\gamma + \sum_{n=1}^{\infty} \frac{a+1}{n (n+a+1)} + \gamma \right ) \, {\rm d}a\\
&= \int_{0}^{1} \sum_{n=1}^{\infty} \frac{1}{n(n+a+1)} \, {\rm d}a\\
&= \sum_{n=1}^{\infty}\frac{1}{n} \int_{0}^{1} \frac{{\rm d}a}{n+a+1} \\
&= \sum_{n=1}^{\infty} \frac{\log(n+2) - \log(n+1)}{n} \\
&=\gamma_1(2, 0) - \gamma_1(1, 0)
\end{align*}
where $\gamma_n (k, j)$ are the generalized poly Stieljes constants.
Well another way of evaluating the integral $\displaystyle \int_0^1 u^a \ln (1-u) \, {\rm d}u$ is by expanding the $\ln (1-u)$ to Taylor series and use the celebrated formula:
Delete$$\int_0^1 \frac{x^a - x^b}{\ln x} {\rm d}x = \ln \left(\frac{a+1}{b+1}\right)$$
At least we are avoiding the digamma crap.