Integral equality
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Is it true that
\[\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx?\]
Is it true that
\[\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx?\]
1 Answer
The answer is true. We have
\[\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx+\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx\]
\[=0+\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx.\]
Hence
\[\int_{-\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=\int_{\frac{\pi}{2}}^{\pi}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx.\]
Note that
\[\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{\sin ^3 (\pi x)}{1+x^{10}}dx=0\]
since the fuction $\frac{\sin ^3 (\pi x)}{1+x^{10}}$ is odd.
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I edited the question.