Integral exercise with limit and summation asked by Dan Sitaru in many math groups

Exercise

Find $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{\int_{0}^{1}{\frac{x\sin \pi x}{x+\left( 1-x \right){{k}^{1-2x}}}dx}}$ 

Solution: Let $I=\int_{0}^{1}{\frac{x\sin \left( \pi x \right)}{x+\left( 1-x \right){{k}^{1-2x}}}dx}$ 

$I=\int_{0}^{1}{\frac{\left( 1-x \right)\sin \pi \left( 1-x \right)}{\left( 1-x \right)+\left( 1-\left( 1-x \right) \right){{k}^{1-2\left( 1-x \right)}}}dx}$

   $=\int_{0}^{1}{\frac{\sin \pi \left( 1-x \right)}{\left( 1-x \right)+x{{k}^{-1+2x}}}dx-\int_{0}^{1}{\frac{x\sin \pi \left( 1-x \right)}{\left( 1-x \right)+x{{k}^{-1+2x}}}dx}}$  , $\sin \left( \pi -\pi x \right)=\sin \pi x$ 

    $=\int_{0}^{1}{\frac{\sin \left( \pi x \right)}{1-x+x{{k}^{-\left( 1-2x \right)}}}dx-\int_{0}^{1}{\frac{x\sin \left( \pi x \right)}{1-x+x{{k}^{-\left( 1-2x \right)}}}dx}}$ 

But $1-x+x{{k}^{-\left( 1-2x \right)}}=\left( 1-x \right)+\frac{x}{{{k}^{\left( 1-2x \right)}}}=\frac{\left( 1-x \right){{k}^{\left( 1-2x \right)}}+x}{{{k}^{\left( 1-2x \right)}}}$ 

So $I=\int_{0}^{1}{\frac{\sin \left( \pi x \right){{k}^{1-2x}}}{x+\left( 1-x \right){{k}^{1-2x}}}dx}-\int_{0}^{1}{\frac{x\sin \left( \pi x \right){{k}^{1-2x}}}{x+\left( 1-x \right){{k}^{1-2x}}}dx}=\int_{0}^{1}{\frac{\left( 1-x \right)\sin \left( \pi x \right){{k}^{1-2x}}}{x+\left( 1-x \right){{k}^{1-2x}}}dx}$ 

Hence \[2I=\int_{0}^{1}{\frac{x+\left( 1-x \right){{k}^{1-2x}}}{x+\left( 1-x \right){{k}^{1-2x}}}\sin \left( \pi x \right)dx}=\int_{0}^{1}{\sin \left( \pi x \right)dx}=-\frac{1}{\pi }\left( \cos \left( \pi x \right) \right)_{0}^{1}=\frac{1}{\pi }\] 

Thus $\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{k=1}^{n}{\frac{1}{\pi }}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\times \frac{1}{\pi }\sum\limits_{k=1}^{n}{1}=\frac{1}{\pi }$ 

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the idea of solution credit to Vulvaal Pal