Answer to Question #309865 in Calculus for Swagatika

Question

If ln=int_zero to infinity e^-x sin^n (x)dx,prove that (1+n^2)ln=n(n-1)ln-2 for n≥2.

Answer 1

Solution

$In = \int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} \$

We solve the given integral by integration by parts method

$u = {e^{ - x}}\$ and $dv = \sin \left( {nx} \right)\,dx\$

$du = - {e^{ - x}}dx\$ and $v = - \frac{{\cos \left( {nx} \right)}}{n}$

Then 

$In = \int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} \$

$In = \left[ { - \frac{{{e^{ - x}}\cos \left( {nx} \right)}}{n}} \right]_0^\infty - \int\limits_0^\infty {\left( { - \frac{{\cos \left( {nx} \right)}}{n}} \right)\left( { - {e^{ - x}}} \right)\,dx}$

$In = - \frac{1}{n}\left[ {\frac{{\cos \left( {nx} \right)}}{{{e^x}}}} \right]_0^\infty - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}$

$In = - \frac{1}{n}\left[ {\frac{{\cos \left( {n\left( \infty \right)} \right)}}{{{e^\infty }}} - \frac{{\cos \left( {n\left( 0 \right)} \right)}}{{{e^{\left( 0 \right)}}}}} \right] - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}$

$In = - \frac{1}{n}\left[ {0 - 1} \right] - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}$

$In = \frac{1}{n} - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx}$

 $\begin{array}{l} In = \frac{1}{n} - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx} \\ \end{array}\$

 $\begin{array}{l} In = \frac{1}{n} - \frac{1}{n}\left( I \right) & & & ... & \left( 1 \right) \end{array}\$

Now consider

$I = \int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx} \$

Solving by integration by parts 

$u = {e^{ - x}}$  and  $dv = \cos \left( {nx} \right)\,dx$

$du = - {e^{ - x}}dx\$  and  $v = \frac{{\sin \left( {nx} \right)}}{n}$

Then

$I = \left[ {\frac{{{e^{ - x}}\sin \left( {nx} \right)}}{n}} \right]_0^\infty - \int\limits_0^\infty {\frac{{ - {e^{ - x}}\sin \left( {nx} \right)}}{n}dx}$

$I = \frac{1}{n}\left[ {\frac{{\sin \left( {nx} \right)}}{{{e^x}}}} \right]_0^\infty + \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)dx}$

$I = \frac{1}{n}\left[ {0 - 0} \right]_0^\infty + \frac{1}{n}\left( {In} \right)$

$I = \frac{1}{n}\left( {In} \right)$

Hence from (1), we have 

$In = \frac{1}{n} - \frac{1}{n}\left( {\frac{1}{n}\left( {In} \right)} \right)\\$

$In = \frac{1}{n} - \frac{1}{{{n^2}}}\left( {In} \right)\\$

$In\left( {1 + \frac{1}{{{n^2}}}} \right) = \frac{1}{n}\\$

$In = \frac{1}{n} \times \frac{{{n^2}}}{{1 + {n^2}}}\\$

$In = \frac{n}{{1 + {n^2}}}$

Therefore,

$\int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} = \frac{n}{{1 + {n^2}}}$

For the given (1+n^2)ln=n(n-1)ln-2 for n≥2, its not clear. However, if a weel typed equation is added, can be proved accordingly.