TITLE: Evaluate ${\int_0^\infty e^{-t} \log(\cos^2 t)}\,\mathrm dt$
QUESTION [15 upvotes]: After reading a textbook about integrals, my curiosity sparks about the following integral:
$$\int_0^\infty e^{-t} \log(\cos^2 t)\, \mathrm dt$$
How to evaluate a closed form of this integral ?
My bet is to use the Cauchy Integral Theorem of Residues or the Laplace Transform.

REPLY [4 votes]: Another approach
Using the formula
$$
\sum_{k=1}^\infty\left(\frac1k-\frac1{k+z}\right)=\psi(z+1)+\gamma\tag{1}
$$
we get
$$
\begin{align}
\sum_{k=1}^\infty(-1)^k\frac1{k+z}
&=\sum_{k=1}^\infty\left(\frac1{2k}-\frac1{2k-1+z}\right)-\left(\frac1{2k}-\frac1{2k+z}\right)\\
&=\frac12\sum_{k=1}^\infty\left(\frac1{k}-\frac1{k+\frac{z-1}{2}}\right)
-\frac12\sum_{k=1}^\infty\left(\frac1{k}-\frac1{k+\frac{z}{2}}\right)\\
&=\frac12\left(\psi\left(\frac{z+1}{2}\right)+\gamma\right)
-\frac12\left(\psi\left(\frac{z+2}{2}\right)+\gamma\right)\\
&=\frac12\psi\left(\frac{z+1}{2}\right)-\frac12\psi\left(\frac{z+2}{2}\right)\tag{2}
\end{align}
$$
Applying $(2)$ to the question yields
$$
\begin{align}
\int_0^\infty\log(\cos^2(t))\,e^{-t}\,\mathrm{d}t
&=2\,\mathrm{Re}\left(\int_0^\infty\log(\cos(t))\,e^{-t}\,\mathrm{d}t\right)\\
&=2\,\mathrm{Re}\left(\int_0^\infty\left(\color{#C00000}{\log(e^{it})}+\log(1+e^{-2it})\color{#00A000}{-\log(2)}\right)\,e^{-t}\,\mathrm{d}t\right)\\
&=\color{#00A000}{-2\log(2)}-2\sum_{k=1}^\infty\mathrm{Re}\left(\int_0^\infty\frac{(-1)^k}{k}e^{-2kit}\,e^{-t}\,\mathrm{d}t\right)\\
&=-2\log(2)-2\sum_{k=1}^\infty\mathrm{Re}\left(\frac{(-1)^k}{k}\frac1{1+2ki}\right)\\
&=-2\log(2)-2\sum_{k=1}^\infty(-1)^k\mathrm{Re}\left(\frac1k-\frac1{k+\frac1{2i}}\right)\\
&=2\sum_{k=1}^\infty(-1)^k\mathrm{Re}\left(\frac1{k+\frac1{2i}}\right)\\
&=\sum_{k=1}^\infty(-1)^k\left(\frac1{k+\frac1{2i}}+\frac1{k-\frac1{2i}}\right)\\
&=\frac12\psi\left(\frac12+\frac i4\right)
+\frac12\psi\left(\frac12-\frac i4\right)\\
&-\frac12\psi\left(1+\frac i4\right)
-\frac12\psi\left(1-\frac i4\right)\tag{3}
\end{align}
$$
This agrees with the other answers, and plugging into Mathematica yields
N[1/2(PolyGamma[1/2+I/4]+PolyGamma[1/2-I/4]
-PolyGamma[1+I/4]-PolyGamma[1-I/4]), 20]
$-1.03225543583966217044$