Integral exercise asked by Hamza Mahmood in their facebook wall

Exercise:

Compute, $\int_{\pi }^{\infty }{\frac{dx}{x+{{x}^{\pi }}}}$

Solution: we have $\frac{1}{x+{{x}^{\pi }}}=\frac{1}{x\left( 1+{{x}^{\pi -1}} \right)}$

Put $x={{e}^{t}}\Leftrightarrow dx={{e}^{t}}dt$

So $\int_{\pi }^{\infty }{\frac{dx}{x+{{x}^{\pi }}}=\int_{\ln \pi }^{\infty }{\frac{{{e}^{t}}dt}{{{e}^{t}}\left( 1+{{\left( {{e}^{t}} \right)}^{\pi -1}} \right)}=\int_{\ln \pi }^{\infty }{\frac{dt}{1+{{e}^{t\left( \pi -1 \right)}}}}}}$

Let $u=1+{{e}^{t\left( \pi -1 \right)}}\Leftrightarrow {{e}^{t\left( \pi -1 \right)}}=u-1\Leftrightarrow du=\left( \pi -1 \right){{e}^{t\left( \pi -1 \right)}}dt\Leftrightarrow du=\left( \pi -1 \right)\left( u-1 \right)dt$

So $\int_{\ln \pi }^{\infty }{\frac{dt}{1+{{e}^{t\left( \pi -1 \right)}}}}=\int_{1+\pi {{e}^{\pi -1}}}^{\infty }{\frac{\frac{du}{\left( \pi -1 \right)\left( u-1 \right)}}{u}}=\frac{1}{\pi -1}\int_{1+\pi {{e}^{\pi -1}}}^{\infty }{\frac{du}{u\left( u-1 \right)}}$

But $\frac{1+u-u}{u\left( u-1 \right)}=\frac{-\left( u-1 \right)+u}{u\left( u-1 \right)}=\frac{-1}{u}+\frac{1}{u-1}$

So $\int_{1+\pi {{e}^{\pi -1}}}^{\infty }{\frac{du}{u\left( u-1 \right)}=\int_{1+\pi {{e}^{\pi -1}}}^{\infty }{\frac{du}{u-1}-\int_{1+\pi {{e}^{\pi -1}}}^{\infty }{\frac{du}{u}}}}=\ln \left| u-1 \right|_{1+\pi {{e}^{\pi -1}}}^{\infty }-\ln \left| u \right|_{1+\pi {{e}^{\pi -1}}}^{\infty }$

Thus $\int_{\pi }^{\infty }{\frac{dx}{x+{{x}^{\pi }}}}=\frac{1}{\pi -1}\int_{1+\pi {{e}^{\pi -1}}}^{\infty }{\frac{du}{u\left( u-1 \right)}}=\frac{1}{\pi -1}\ln \left| \frac{u-1}{u} \right|_{1+\pi {{e}^{\pi -1}}}^{\infty }=\frac{1}{\pi -1}\ln \left| \frac{{{x}^{\pi -1}}}{1+{{x}^{\pi -1}}} \right|_{\pi }^{\infty }$

$=\frac{1}{\pi -1}\ln \left| \frac{{{\pi }^{\pi -1}}+1}{{{\pi }^{\pi -1}}} \right|=\frac{1}{\pi -1}\ln \left| \frac{{{\pi }^{\pi }}+\pi }{{{\pi }^{\pi }}} \right|$

Or: we can start as follows :

We have $\int_{\pi }^{\infty }{\frac{dx}{x+{{x}^{\pi }}}=\int_{\pi }^{\infty }{\frac{dx}{{{x}^{\pi }}\left( 1+{{x}^{1-\pi }} \right)}=}\int_{\pi }^{\infty }{\frac{{{x}^{-\pi }}dx}{1+{{x}^{1-\pi }}}}}$

Let $u=1+{{x}^{1-\pi }}\Leftrightarrow du=\left( 1-\pi  \right){{x}^{-\pi }}dx\Leftrightarrow {{x}^{-\pi }}dx=\frac{du}{1-\pi }$

But $u\left( \theta  \right)=1+{{\theta }^{1-\pi }}\Leftrightarrow \underset{\theta \to \infty }{\mathop{\lim }}\,1+\frac{\theta }{{{\theta }^{\pi }}}=1+\underset{\theta \to \infty }{\mathop{\lim }}\,\frac{1}{\pi {{\theta }^{\pi -1}}}=1$ and $u\left( \pi  \right)=1+{{\pi }^{1-\pi }}$

So $\int_{\pi }^{\infty }{\frac{{{x}^{-\pi }}dx}{1+{{x}^{1-\pi }}}=\frac{1}{1-\pi }\underset{\theta \to \infty }{\mathop{\lim }}\,\int_{u\left( \pi  \right)}^{u\left( \theta  \right)}{\frac{du}{u}}}=\frac{1}{1-\pi }\underset{\theta \to \infty }{\mathop{\lim }}\,\ln \left| u \right|_{u\left( \pi  \right)}^{u\left( \theta  \right)}=\frac{1}{1-\pi }\ln \left| \frac{1}{1+{{\pi }^{1-\pi }}} \right|$