2022-07-01

Proving ${\int }_{0}^{1}\frac{\mathrm{sin}\pi x}{{x}^{2}+1}dx=\frac{2}{\pi \left({\mu }_{1}^{2}+1\right)}=\frac{\pi }{4}\mathrm{sin}\pi {\mu }_{2}$ for certain ${\mu }_{1},{\mu }_{2}\in \left[0,1\right]$

Hint: Notice that $\underset{0}{\overset{1}{\int }}\mathrm{sin}\pi t\phantom{\rule{thinmathspace}{0ex}}dt=\frac{2}{\pi }$ and $\underset{0}{\overset{1}{\int }}\frac{1}{1+{t}^{2}}\phantom{\rule{thinmathspace}{0ex}}dt=\frac{\pi }{4}$. The generalized mean value theorem states that for f,g continuous on an interval [a,b] and differentiable on (a,b), there exists $c\in \left(a,b\right)$ such that
$\left(f\left(b\right)-f\left(a\right)\right){g}^{\prime }\left(c\right)=\left(g\left(b\right)-g\left(a\right)\right){f}^{\prime }\left(c\right).$
Apply this theorem for
$f\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{\mathrm{sin}\pi t}{{t}^{2}+1}\phantom{\rule{thinmathspace}{0ex}}dt$
and
$g\left(x\right)=\underset{0}{\overset{x}{\int }}\mathrm{sin}\pi t\phantom{\rule{thinmathspace}{0ex}}dt\phantom{\rule{1cm}{0ex}}\text{or}\phantom{\rule{1cm}{0ex}}g\left(x\right)=\underset{0}{\overset{x}{\int }}\frac{1}{{t}^{2}+1}\phantom{\rule{thinmathspace}{0ex}}dt.$

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