Integrals at infinity/ factorials problem

[Samme013]

View attachment 3440
Need help on exercise 2 from the linked image , left first in so you guys could see the Γ(χ) function any help is appreciated , thanks in advance!

 

[Siron]

2.(a) Apply the substitution $t=u^2$ for $\Gamma(x)$.
2.(b) Can you show where you're stuck?

 

[Samme013]

2.(a) Apply the substitution $t=u^2$ for $\Gamma(x)$.
2.(b) Can you show where you're stuck?

Ok i did a and n b i proved it for n=0 assumed for n=k and used the fact that
Γ(x+1)=xΓ(χ) when calculating Γ(χ+1+ 1/2) but could not find a way το prove that it is equal with what one gets just by plugging n=k+1 in the statement that we want to prove

 

[Euge]

Ok i did a and n b i proved it for n=0 assumed for n=k and used the fact that
Γ(x+1)=xΓ(χ) when calculating Γ(χ+1+ 1/2) but could not find a way το prove that it is equal with what one gets just by plugging n=k+1 in the statement that we want to prove

Use 1(b) to write

\(\displaystyle \Gamma\Bigl((k + 1) + \frac{1}{2}\Bigr) = \Gamma\left(\Bigl(k + \frac{1}{2}\Bigr) + 1\right) = \Bigl(k + \frac{1}{2}\Bigr) \Gamma\Bigl(k + \frac{1}{2}\Bigr),\)

and then use the induction hypothesis

\(\displaystyle \Gamma\Bigl(k + \frac{1}{2}\Bigr) = \frac{(2k)!}{4^k k!}\sqrt{\pi}.\)

 

[Samme013]

Use 1(b) to write

\(\displaystyle \Gamma\Bigl((k + 1) + \frac{1}{2}\Bigr) = \Gamma\left(\Bigl(k + \frac{1}{2}\Bigr) + 1\right) = \Bigl(k + \frac{1}{2}\Bigr) \Gamma\Bigl(k + \frac{1}{2}\Bigr),\)

and then use the induction hypothesis

\(\displaystyle \Gamma\Bigl(k + \frac{1}{2}\Bigr) = \frac{(2k)!}{4^k k!}\sqrt{\pi}.\)

Yes that is as far as i go but how is that equal to the statement for n=k+1

 

[Euge]

Yes that is as far as i go but how is that equal to the statement for n=k+1

I gave you a hint to solve the problem. You now have to show

\(\displaystyle \Bigl(k + \frac{1}{2}\Bigr)\frac{(2k)!}{4^k k!}\sqrt{\pi} = \frac{(2k+2)!}{4^{k+1} (k+1)!}\sqrt{\pi}.\)

To do so, write $k + \frac{1}{2} = \frac{2k + 1}{2}$ and use the identity

\(\displaystyle \frac{1}{2} = \frac{2k + 2}{4(k + 1)}.\)

 

[Samme013]

I gave you a hint to solve the problem. You now have to show

\(\displaystyle \Bigl(k + \frac{1}{2}\Bigr)\frac{(2k)!}{4^k k!}\sqrt{\pi} = \frac{(2k+2)!}{4^{k+1} (k+1)!}\sqrt{\pi}.\)

To do so, write $k + \frac{1}{2} = \frac{2k + 1}{2}$ and use the identity

\(\displaystyle \frac{1}{2} = \frac{2k + 2}{4(k + 1)}.\)

Wow that was kinda simple , i def need more practice manipulating factorials , thanks for the help man.