$$
\boxed{
\frac {1}\pi \int _{0}^{1}\frac {x^{x}(1-x)^{1-x}\sin \pi x}{1-ax}dx
=\frac {1}{a^{2}}\left (e-\frac {ea}2-(1-a)^{1-1/a}\right )
}
$$
証明
$$
\begin {aligned}
&自然複素対数関数\ln _{0}z,\ln _{1}zの偏角を、\\
&\quad \begin {aligned}
0&\leq \Im \ln _{0}z<2\pi ,\\
-\pi &\leq \Im \ln _{1}z<\pi
\end{aligned}\\
&となるようにとり、\\
&\quad \begin {aligned}f(z)&=\exp \left (\pi iz+z\ln _{1}z+(1-z)\ln _{0}(1-z)\right )/(1-az)\end {aligned}\\
&とおく。\\
&f(z)は\mathbb C \setminus\left ([0,1]\cup\left \{1/a\right \}\right )で正則で、\\
&\quad \begin {aligned}
\oint _{z=re^{i\theta }}f(z)dz&\to 0\quad (r\to 0),また\\
\oint _{z=1+re^{i\theta }}f(z)dz&\to 0\quad (r\to 0)
\end {aligned}\\
&なので、z=1/aを内部に含まない、[0,1]を囲うような積分経路Cについて、\\
&\quad \begin {aligned}
\int _{C}f(z)dz&=\int _{0}^{1}f(x-0i)dx+\int _{1}^{0}f(x+0i)dx\\
&=\int _{0}^{1}\left (\exp \left (\pi ix+x\ln x+(1-x)\ln x\right )-\exp \left (\pi ix+x\ln x+(1-x)\left (\ln (1-x)+2\pi i\right )\right )\right )/(1-ax)dx\\
&=\int _{0}^{1}\left (\exp (\pi ix)-\exp (-\pi ix)\right )\exp \left (x\ln x+(1-x)\ln (1-x)\right )/(1-ax)dx\\
&=2i\int _{0}^{1}\frac {\sin (\pi x)x^{x}(1-x)^{1-x}}{1-ax}dx
\end {aligned}\\
&一方で,C'=\{z|z^{-1}\in C\}とすると\\
&\quad \begin {aligned}
\int _Cf(z)dz
&=-\int _{C'}f\left (\frac {1}z\right )\frac {dz}{z^2}\end {aligned}\\
&である。x\in (0,1)\setminus \{a\}のとき\\
&\quad \begin {aligned}
f\left (\frac {1}x\right )&=\frac {1}{1-\frac {a}x}\exp\left (\frac {\pi i}x-\frac {\ln x}x+\frac {x-1}x\left (\ln \frac {x-1}x+\pi i\right )\right )\\
&=\frac {x}{x-a}\exp\left (\frac {1}x\left (\pi i-\ln x-\ln \frac {x-1}x-\pi i\right )+\ln (x-1)-\ln x+\pi i\right )\\
&=\frac {x}{a-x}\exp \left (-\ln x+\left (1-\frac {1}x\right )\ln (1-x)\right )\\
&=\frac {(1-x)^{1-1/x}}{a-x}
\end {aligned}\\
&であり、C'の内の\frac {1}{z^{2}}f\left (\frac {1}z\right )の極は\\
&z=0に2位の極で留数は\\
&\quad \begin {aligned}
\underset{z=0}{\mathrm{Res}}\frac {1}{z^{2}}f\left (\frac {1}z\right )&=\lim _{z\to 0}\frac {d}{dz}f\left (\frac {1}z\right )\\
&=\lim _{z\to 0}\frac {(a-z)(\frac {d}{dz}(1-z)^{1-1/z})+(1-z)^{1-1/z}}{(a-z)^{2}}\\
&=\lim _{z\to 0}\frac {1}{a^{2}}\left (1-z\right )^{1-1/z}\left (1+a\cdot \frac {d}{dz}\left (\left (1-\frac {1}z\right )\ln (1-z)\right )\right )\\
&=\lim _{z\to 0}\frac {e}{a^{2}}\left (1+a\left (\frac {z-1}z\cdot \frac {1}{z-1}+\frac {\ln (1-z)}{z^{2}}\right )\right )\\
&=\lim _{z\to 0}\frac {e}{a^{2}}\left (1+a\left (\frac {z+\ln (1-z)}{z^{2}}\right )\right )\\
&=\lim _{z\to 0}\frac {e}{a^{2}}\left (1+a\cdot \frac {1-\frac {1}{1-z}}{2z}\right )\\
&=\frac {e}{a^{2}}\left (1-\frac {a}2\right )
\end {aligned}\\
&z=aに1位の極で留数は\\
&\quad \begin {aligned}
\underset{z=a}{\mathrm{Res}}\frac {1}{z^{2}}f\left (\frac {1}z\right )
&=\lim _{z\to a}\frac {z-a}{z^{2}}\cdot \frac {(1-z)^{1-1/z}}{a-z}\\
&=-\frac {\left (1-a\right )^{1-1/a}}{a^2}
\end {aligned}\\
&従って\\
&\quad \begin {aligned}
-\int _{C'}\frac {1}{z^{2}}f\left (\frac {1}z\right )dz&=2\pi i\left (\frac {e}{a^{2}}\left (1-\frac {a}2\right )-\frac {(1-a)^{1-1/a}}{a^{2}}\right )\\
\frac {1}\pi \int _{0}^{1}\frac {x^{x}(1-x)^{1-x}\sin \pi x}{1-ax}dx
&=\textcolor {blue}{\frac {1}{a^{2}}\left (e-\frac {ea}2-(1-a)^{1-1/a}\right )}.
\end {aligned}
\end {aligned}
$$