By parts integrations applications

Exercise:

Consider the sequence ${{\left( {{I}_{n}} \right)}_{n>0}}$ is defined to be ${{I}_{n}}=\int_{1}^{e}{x{{\left( \ln x \right)}^{n}}dx}$

Show that, ${{e}^{2}}=2{{I}_{n}}+n{{I}_{n-1}}$ for every $n\in \mathbb{N}$

Solution: We have${{I}_{n}}=\int_{1}^{e}{x{{\left( \ln x \right)}^{n}}dx}$

Let $u={{\left( \ln x \right)}^{n}}\And dv=x\Leftrightarrow du=\frac{n}{x}{{\left( \ln x \right)}^{n-1}}\And v=\frac{{{x}^{2}}}{2}$ , then

${{I}_{n}}=\int_{1}^{e}{x{{\left( \ln x \right)}^{n}}dx}=\left[ \frac{{{x}^{2}}}{2}{{\left( \ln x \right)}^{n}} \right]_{1}^{e}-\int_{1}^{e}{\frac{{{x}^{2}}}{2}\frac{n}{x}{{\left( \ln x \right)}^{n-1}}dx}=\frac{{{e}^{2}}}{2}-\frac{n}{2}\int_{1}^{e}{x{{\left( \ln x \right)}^{n-1}}dx}$

Hence, $2{{I}_{n}}={{e}^{2}}-n{{I}_{n-1}}\Leftrightarrow {{e}^{2}}=2{{I}_{n}}-n{{I}_{n-1}}$                                  Q.E.D



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Exercise:

Consider the sequence ${{\left( {{J}_{n}} \right)}_{n}}$ defined to be ${{J}_{n}}=\int_{0}^{\pi /2}{{{\sin }^{n}}x}\,dx$

Show that, for all natural numbers $n$ we have $\left( n+2 \right){{J}_{n+2}}=\left( n+1 \right){{J}_{n}}$

Solution: we have ${{J}_{n+2}}=\int_{0}^{\pi /2}{{{\sin }^{n+2}}x\,dx}=\int_{0}^{\pi /2}{{{\sin }^{n+1+1}}xdx}=\int_{0}^{\pi /2}{{{\sin }^{n+1}}x\sin xdx}$

Let $u={{\sin }^{n+1}}x\And dv=\sin x\Leftrightarrow du=\left( n+1 \right){{\sin }^{n}}x\cos xdx\And v=-\cos x$

${{J}_{n+2}}=\int_{0}^{\pi /2}{{{\sin }^{n+2}}dx}=-\left( {{\sin }^{n+1}}x\cos x \right)_{0}^{\pi /2}+\int_{0}^{\pi /2}{\cos x\left( n+1 \right){{\sin }^{n}}x\cos xdx}$

       $=\left( n+1 \right)\int_{0}^{\pi /2}{{{\sin }^{n}}x{{\cos }^{2}}xdx}=\left( n+1 \right)\int_{0}^{\pi /2}{{{\sin }^{n}}x\left( 1-{{\sin }^{2}}x \right)dx}$

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

       $=\left( n+1 \right){{J}_{n}}-\left( n+1 \right){{J}_{n+2}}$

Thus, $\left( n+1+1 \right){{J}_{n+2}}=\left( n+1 \right){{J}_{n}}\Leftrightarrow \left( n+2 \right){{J}_{n+2}}=\left( n+1 \right){{J}_{n}}$