How to integrate $\int_{0}^{1}\ln\left(\, x\,\right)\,{\rm d}x$?

Question

I encountered this integral in the quantum field theory calculation. Can I do this:

$$ \left. \int_{0}^{1}\ln\left(\, x\,\right)\,{\rm d}x =x\ln\left(\, x\,\right)\right\vert_{0}^{1} -\int_{0}^{1}\,{\rm d}x =\left. x\ln\left(\, x\,\right)\right\vert_{\, x\ =\ 0}\ -\ 1 $$

So the first term looks divergent. But Mathematica gives finite result and the integral is $-1$. Why isn't the first term divergent ?.

Answer 1

Use L'Hopital's Rule to resolve the indeterminate form: $$\lim_{x\to 0^+}x\ln x=\lim_{x\to 0^+}{\ln x\over x^{-1}}=\lim_{x\to 0^+}{x^{-1}\over -x^{-2}}=\lim_{x\to 0^+}(-x)=0$$

Answer 2

Hint: Use L'Hopital's rule to evaluate $\lim_{x\to0^+}x\ln x$.

Answer 3

Or alternatively, use the substitution $x=e^y$.

Answer 4

interpret $\int_0^1 \ln x dx$ as the signed area of the region bounded by $x = 0, y= \ln(x)$ and $y = 0$ this is the negative of the area of the region bounded by $x = e^y, y = 0$ and $x = 0$(interchange $x$ and $y$ in the previous boundary). the area of latter region is $\int_{-\infty}^0 e^y dy = 1.$ so $\int_0^1 \ln(x) dx = -1.$

another way to think of the same thing is in the first area computation you have the infinitesimal rectangles parallel to the $y$-axis in the latter the infinitesimal rectangles parallel to $x$-axis.

Answer 5

We have $$\begin{gathered} \int\limits_0^1 {\ln xdx} = \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \int\limits_\varepsilon ^1 {\ln xdx} = \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \left( {\left. {\left( {x\ln x} \right)} \right|_{x = \varepsilon }^{x = 1} - \int\limits_\varepsilon ^1 {dx} } \right) = \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \left( { - \varepsilon \ln \varepsilon - 1 + \varepsilon } \right) \hfill \\ = - \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \varepsilon \ln \varepsilon - 1. \hfill \\ \end{gathered}$$ On the other hand, by the L'Hospital rule, we have $$\mathop {\lim }\limits_{\varepsilon \to {0^ + }} \varepsilon \ln \varepsilon = \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \frac{{\ln \varepsilon }} {{\frac{1} {\varepsilon }}} = \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \frac{{\frac{1} {\varepsilon }}} {{ - \frac{1} {{{\varepsilon ^2}}}}} = \mathop {\lim }\limits_{\varepsilon \to {0^ + }} \left( { - \varepsilon } \right) = 0.$$ So $$\int\limits_0^1 {\ln xdx} = - 1.$$

Answer 6

Consider the general log integral, $$I_n=\int_0^1\log^n(x)\ dx$$ note that we are after $I_1$ and define the exponential generating function $G(t)$, $$G(t)=\sum_{n=0}^\infty\frac{I_n}{n!}t^n=\int_0^1\sum_{n=0}^\infty\frac{(t\log(x))^n}{n!}\ dx=\int_0^1x^t\ dx=\frac{1}{1+t}$$ for $t>-1$ and expanding the result as a geometric series in the same region, $$\frac{1}{1+t}=\sum_{n=0}^\infty(-1)^nt^n=\sum_{n=0}^\infty\frac{(-1)^nn!}{n!}t^n$$ comparing coefficients, $$I_n=(-1)^nn!.$$ The first four non-zero log integrals are, $$\begin{align*} &\int_0^1\log(x)\ dx=-1,\quad\int_0^1\log^2(x)\ dx=2 \\ &\int_0^1\log^3(x)\ dx=-6,\quad\int_0^1\log^4(x)\ dx=24. \end{align*}$$