Summation of Integrals Mashup!!

Calculus Level 4

$Let\quad { I }_{ n }=\quad\displaystyle \int _{ 0 }^{ \pi /2 }{ { x }^{ n }\cos { x } dx,\quad where\quad n\quad is\quad a\quad non-negative\quad integer. } \\ Then:\quad \quad \displaystyle\sum _{ n=2 }^{ \infty }{ \left( \frac { { I }_{ n } }{ n! } +\frac { { I }_{ n-2 } }{ (n-2)! } \right) } equals??$
Also Try: $\href{https://brilliant.org/community-problem/integrals-of-the-years/?group=mJXtqmVrO6YT&ref_id=444794}{here}$

The answer is 2.239.

1 solution

Arghyadeep Chatterjee
Oct 17, 2019

The reduction formula can be easily derived for this integral which is

$\large I_{n}=(\frac{\pi}{2})^{n} - n(n-1)I_{n-2}$

So dividing by $n!$ and rearranging we have

$\large \frac{I_{n}}{n!}+\frac{I_{n-2}}{(n-2)!} =\frac{(\frac{\pi}{2})^{n}}{n!}$

So using Maclaurin's expansion for $e^x$ we have

$\large e^x=\sum_{r=0}^{\infty} \frac{x^r}{r!}$

So putting $\large x=\frac{\pi}{2}$ we have :-

$\large \sum_{r=0}^{\infty} \frac{(\frac{\pi}{2})^{r}}{r!} = e^{\frac{\pi}{2}}$

So $\large \displaystyle\sum_{r=2}^{\infty} \frac{(\frac{\pi}{2})^{r}}{r!} = e^{\frac{\pi}{2}}-1-\frac{\pi}{2}$

So our answer is

$\large \displaystyle e^{\frac{\pi}{2}}-1-\frac{\pi}{2}$

Summation of Integrals Mashup!!

The answer is 2.239.

1 solution

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