TITLE: What is $\int_{0}^{\infty}\frac{\cos^{n}x \sin x \ln x}{x} dx$?
QUESTION [6 upvotes]: Letting $\gamma$ denote the Euler-Mascheroni constant, evaluating $\int_{0}^{\infty}\frac{\cos^{n}x \sin x \ln x}{x} dx$ for $n \in \{ 1, 2, 3, 4 \}$, we have that: 
$$\begin{align*}
 \int_{0}^{\infty}\frac{\cos^{1}x \sin x \ln x}{x} dx & = - \frac{1}{4} \pi  \left(  \gamma  + \ln 2  \right),   \\
 \int_{0}^{\infty}\frac{\cos^{2}x \sin x \ln x}{x} dx & = - \frac{1}{8} \pi  \left(  2  \gamma  + \ln 3  \right),   \\
 \int_{0}^{\infty}\frac{\cos^{3}x \sin x \ln x}{x} dx & = - \frac{1}{16} \pi  \left(  3  \gamma  + \ln 16  \right),   \\
 \int_{0}^{\infty}\frac{\cos^{4}x \sin x \ln x}{x} dx & = - \frac{1}{32} \pi  \left(  6  \gamma  + \ln 135  \right).
\end{align*}
$$
More generally, it appears that $$\int_{0}^{\infty}\frac{\cos^{n}x \sin x \ln x}{x} dx = - \frac{1}{2^{n+1}} \pi \left( \binom{n}{ \left\lfloor \frac{n}{2} \right\rfloor  } \gamma + \ln a_{n} \right)$$ for some integer sequence $$(a_{n} : n \in \mathbb{N}) = \left( 2, 3, 16, 135, 49152, 430565625, 1641562064176545792, \ldots  \right)$$ which is not currently in the On-Line Encyclopedia of Integer Sequences.
Ideally, I would like to find a closed-form evaluation for this sequence. Alternatively, I would be interested in evaluating this sequence as a summation (e.g., a binomial sum), the terms of which have a closed-form evaluation.
I have tried to evaluate this sequence by evaluating indefinite integrals of the form $\int \frac{\cos^{n}x \sin x \ln x}{x} dx$, but this is difficult since Mathematica is not able to evaluate a general indefinite integral of this form, and Mathematica is only able to evaluate indefinite integrals of this form for specific values of $n \in \mathbb{N}$ using generalized hypergeometric functions and the incomplete gamma function.

REPLY [11 votes]: A plan:

We may write $\cos^n(x)\sin(x)$ as a Fourier sine series, so the problem boils down to evaluating $\int_{0}^{+\infty}\sin(kx)\frac{\log x}{x}\,dx $;
The last integral is $f'(1)$, where $f(a)=\int_{0}^{+\infty}\frac{\sin(kx)}{x^a}\,dx=k^{a-1}\int_{0}^{+\infty}\frac{\sin x}{x^a}\,dx$;
The integral $I(a)=\int_{0}^{+\infty}\frac{\sin x}{x^a}\,dx$ can be easily evaluated through the Laplace transform, leading to:
$$ I(a) = \frac{1}{\Gamma(a)}\int_{0}^{+\infty}\frac{s^{a-1}}{s^2+1}\,ds = \frac{\pi}{2\,\Gamma(a)\sin\left(\frac{\pi a}{2}\right)};$$
By putting all together, we get:

$$ \int_{0}^{+\infty}\sin(kx)\frac{\log x}{x}\,dx = \frac{\pi}{2}\left(\gamma+\log k\right). $$



Now we just need to compute the Fourier sine series of $\cos^n(x)\sin(x)$.