Too many Gammas coming in!

Calculus Level 5

$\large \sum_{r=0}^{\infty} \left(\dfrac{-1}{2}\right)^r \dfrac{\Gamma \left(\frac{1}{2}\right) \Gamma \left(\frac{1+r}{2}\right)}{\Gamma\left(1+\frac{r}{2}\right)}=\dfrac{p_1^{a} \pi^{b}}{p_2^{{c/ }{d}}}$

The above equation holds true where $p_1$ and $p_2$ are primes with integers $a,b,c$ and $d$ such that $c,d$ is coprime.

Find $p_1+p_2+a+b+c+d$ .

The answer is 13.

1 solution

Aditya Kumar
Jul 5, 2016

$S=\sum _{ r=0 }^{ \infty }{ \left[ { \left( \frac { -1 }{ 2 } \right) }^{ r }\frac { \Gamma \left( \frac { 1 }{ 2 } \right) \Gamma \left( \frac { 1+r }{ 2 } \right) }{ \Gamma \left( 1+\frac { r }{ 2 } \right) } \right] }$

Using beta function , we get:

$S=\sum _{ r=0 }^{ \infty }{ \left[ { \left( \frac { -1 }{ 2 } \right) }^{ r }\int _{ 0 }^{ 1 }{ \frac { { \left( \sqrt { t } \right) }^{ r } }{ \sqrt { t } \sqrt { 1-t } } dt } \right] }$

On integer-changing summation and integral, we get:

$S=\int _{ 0 }^{ 1 }{ \left[ \sum _{ r=0 }^{ \infty }{ { \left( \frac { -1 }{ 2 } \right) }^{ r }\frac { { \left( \sqrt { t } \right) }^{ r } }{ \sqrt { t } \sqrt { 1-t } } } \right] dt }$

The summation is an infinite geometric sum.

$S=\int _{ 0 }^{ 1 }{ \left[ \frac { 2 }{ \left( 2+\sqrt { t } \right) \sqrt { t } \sqrt { 1-t } } \right] dt }$

The indefinite integral can be evaluated using $t=x^2$ and simplifying. Therefore,

$S=\frac { -4 }{ \sqrt { 3 } } { \left\{ \arctan { \left( \frac { -2x-1 }{ \sqrt { 3-3{ x }^{ 2 } } } \right) } \right\} }_{ 0 }^{ 1 }=\huge\boxed{\frac { { 2 }^{ 2 }{ \pi }^{ 1 } }{ { 3 }^{ \frac { 3 }{ 2 } } } }$

Note: Fubini's theorem states that if $\displaystyle \int { \sum { |{ f }_{ n }(x)|dx } } <\infty$ and $\displaystyle \sum { \int { |{ f }_{ n }(x)|dx } } <\infty$ , then $\displaystyle \int { \sum { |{ f }_{ n }(x)|dx } }= \sum { \int { |{ f }_{ n }(x)|dx } }$ . In this question both the cases are less then infinity. Hence the inter-changing can be justified.

Moderator note:

As always, interchanging the order limits (in this case summation and integration) has to be justified. Calling it "integer-changing" doesn't remove this restriction :p

Nice... +1. I'd used the trigonometric form of the beta function (pretty much same as what you did).

A Former Brilliant Member - 4 years, 11 months ago

As always, interchanging the order limits (in this case summation and integration) has to be justified. Calling it "integer-changing" doesn't remove this restriction :p

I almost missed out the "the summation is an infinite geometric sum" till you mentioned it lol.

Calvin Lin Staff - 4 years, 11 months ago

I've added the justification as a note. And yes people tend to miss out infinite geometric sum in such problems as it is kinda unexpected to have that.

Aditya Kumar - 4 years, 11 months ago

The condition for fubini's theorem is that $\int \sum |f_n(x)| \, dx < \infty$ . The absolute sign is important, and ignoring it can lead to strange consequences (like when reordering the terms in $\sum \frac{ (-1)^i}{i}$ ).