Integral with log and trigonometric

Prove that

$$\int_0^{\infty} \frac{\log t (1-\cos t)}{t^2}\, {\rm d}t = \frac{\pi}{2} (1-\gamma)$$

where $\gamma$ stands for the Euler - Mascheroni constant.

Solution

We are basing the calculation on a powerful theorem that connects Laplace and Inverse Laplace. It is known that the inverse Laplace of $\frac{\log t}{t}$ is $-\log x - \gamma$ and we are invoking the Laplace of $\frac{1-\cos t}{t}$.

Lemma: The Laplace transform of $\frac{1-\cos t}{t}$ is $\frac{1}{2} \log \left(\frac{1}{s^2}+1 \right)$.

Proof:  The Laplace transform is given by the formula:

$$F(s)= \mathcal{L} \left \{ f(t) \right \} = \int_{0}^{\infty} e^{-st} f(t) \, {\rm d}t$$

Thus the required Laplace will be given by the integral:

$$F(s) = \mathcal{L}\left \{ \frac{1-\cos t}{t} \right \} = \int_{0}^{\infty} e^{-st} \cdot \frac{1- \cos t}{t} \, {\rm d}t  \tag{1} \label{*}$$

Differentiating \eqref{*} with respect to $s$ once we have that:

\begin{align*}
F'(s) &=\frac{\mathrm{d} }{\mathrm{d} s} \int_{0}^{\infty} e^{-st} \cdot \frac{1- \cos t}{t} \, {\rm d}t \\
 &= \int_{0}^{\infty} \frac{\partial }{\partial s} e^{-s t} \cdot \frac{1-\cos t}{t} \, {\rm d}t\\
 &= -\int_{0}^{\infty} e^{-st} \left ( 1- \cos t \right ) \, {\rm d}t\\
 &= -\int_{0}^{\infty} e^{-st} \, {\rm d}t + \int_{0}^{\infty} e^{-st} \cos t \, {\rm d}t \\
 &= -\frac{1}{s} + \frac{s}{s^2+1}
\end{align*}

Integrating back we have:

\begin{align*}
\frac{1}{2}\log \left ( s^2+1 \right ) - \log s  &= \frac{1}{2} \log \left ( s^2+1 \right ) - \frac{1}{2} \log s^2 \\
 &= \frac{1}{2} \log \left ( \frac{s^2+1}{s^2} \right )\\
 &= \frac{1}{2} \log \left ( \frac{1}{s^2}+1 \right )
\end{align*}

Thus $\displaystyle F(s)=\frac{1}{2} \log \left ( \frac{1}{s^2}+1 \right ) + c$. Now for $s=1$ we have that $\displaystyle F(1)=\frac{\log 2}{2}$ (which is pretty much trivial) thus $c=0$ and this proves the given transform.

Now back to our problem we have successively:

\begin{align*}
\int_{0}^{\infty} \frac{\log t \left ( 1-\cos t \right )}{t^2} \, {\rm d}t &= \int_{0}^{\infty} \frac{\log t \left ( 1-\cos t \right )}{t \cdot t} \, {\rm d}t \\
&= -\frac{1}{2}\int_{0}^{\infty} \left ( \log x + \gamma \right ) \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x \\
&=- \frac{1}{2} \int_{0}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \frac{1}{2} \gamma \int_{0}^{\infty} \log \left ( \frac{1}{x^2} +1 \right ) \, {\rm d}x \\
&= \frac{\pi}{2} - \frac{\pi \gamma}{2}\\
&= \frac{\pi}{2} \left ( 1-\gamma \right )
\end{align*}

since

\begin{align*}
\int_{0}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x &= \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x + \int_{1}^{\infty} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x\\
&\!\!\!\!\!\overset{u=1/x}{=\! =\! =\! =\! =\!} \int_{0}^{1}\log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \int_{1}^{0} \log \left ( \frac{1}{x} \right ) \log \left ( x^2+1 \right ) \frac{1}{x^2} \, {\rm d}x \\
&=\int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x - \int_{0}^{1} \frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x \\
&=-2\mathcal{G} - \frac{\pi}{2} - \log 2 - \left ( -2\mathcal{G} + \frac{\pi}{2} - \log 2\right )\\
&=- \pi \cancel{- 2\mathcal{G} - \log 2 + 2\mathcal{G} + \log 2} \\
&= -\pi
\end{align*}

Now we have to compute three integrals, namely,

$$\int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x \; , \; \int_{0}^{1}\frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x$$

and the integral $\displaystyle  \int_{0}^{\infty} \log \left ( \frac{1}{x^2} +1 \right ) \, {\rm d}x$.

Let us begin with the first two.

• The integral $\displaystyle \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x$. We have successively:
  
  \begin{align*}
  \int_{0}^{1} \log x \log \left ( \frac{1}{x^2}+1 \right ) \, {\rm d}x &= \int_{0}^{1} \log x \log \left ( \frac{x^2+1}{x^2} \right ) \, {\rm d}x \\
  &= \int_{0}^{1} \log x \left [ \log \left ( x^2+1 \right ) - 2 \log x \right ] \, {\rm d}x\\
  &=\int_{0}^{1} \log x \log \left ( x^2+1 \right ) \, {\rm d}x - 2 \int_{0}^{1} \log^2 x \, {\rm d}x \\
  &= \int_{0}^{1} \log x \log \left ( x^2+1 \right ) \, {\rm d}x -4 \\
  &= \int_{0}^{1} \log x \sum_{n=1}^{\infty} \frac{(-1)^{n-1} x^{2n}}{n} \, {\rm d}x -4 \\
  &=\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{1}x^{2n} \log x \, {\rm d}x - 4 \\
  &=- \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \left ( 2n+1 \right )^2} - 4 \\
  &= - \sum_{n=1}^{\infty} (-1)^{n-1} \left [ \frac{1}{n} - \frac{2}{2n+1} - \frac{2}{(2n+1)^2} \right ] -4 \\
  &= -2\mathcal{G} - \frac{\pi}{2} - \ln 2 + 4 -4 \\
  &=  -2\mathcal{G} - \frac{\pi}{2} - \ln 2
  \end{align*}
  
  where we used the Catalan's constant defintion $\displaystyle \mathcal{G}=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)^2}$, the series representation of $\ln 2$ namely $\displaystyle \ln 2 = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$ and finally the Liebniz series of $\displaystyle \frac{\pi}{4}=\sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}$.



• The integral $\displaystyle \int_{0}^{1}\frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x$ can be done in a similar manner and the result is:
  
  $$\displaystyle \int_{0}^{1}\frac{\log x \log \left ( 1+x^2 \right )}{x^2} \, {\rm d}x= -2\mathcal{G} + \frac{\pi}{2} - \log 2 $$



• Finally for the integral $\displaystyle \int_0^{\infty} \log \left ( \frac{1}{x^2}+1 \right )\, {\rm d}x$ we apply integration by parts and we get that its value is actually $\pi$.