Solving the Gamma Function: Using Recursion & Tables

[Erbil]

Homework Statement

Questions are in picture.

Homework Equations

$$ \int _{0}^{\infty }x^{n}e^{-x}dx $$ = $$ Gamma (n+1) = n!
$$ Gamma(P+1) $$ = $$Gamma(P)$$
$$ Gamma(P) = (1/P) $$Gamma(P+1)$$

The Attempt at a Solution

2) I have found it from table.
3) I have used recursion and table to find it.
4) Again With recursion.
5) $$ \Gamma(0.7) $$ = 1/p(p+1) with this formula.
8) If $$ \ Gamma (p+1) $$ is equal to this integral,I think it can be written as $$ \ Gamma (2/3+1) $$ later we can found the value from table.Am I right?
Same logic again for 9,10?
But what next? How can I convert them to Gamma function?

 

[Erbil]

11) I found 1/2 but I'm not sure(?)

Edit : No it's not true.I think true solution is on down.

 

[Erbil]

$$ \int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3 $$

 

[haruspex]

$$ \int _{0}^{\infty }x^{2}e^{-x^{2}}dx=2\int _{0}^{\infty }u^{2}e^{-u}du=2\cdot \Gamma 3 $$

No. How did you get that?
What did you get for (5)?
I don't follow your logic for (8). Where did 2/3 come from? I suspect what you posted is not what you meant.

 

[Erbil]

5. 1/0.7(Gamma1.7) = 1.30 ? Isn't it?
I'm sorry questions mess up.It's solution of question 11.
Here is for (8)

For n= 2/3 Gamma(n+1) = Gamma(1+2/3)=Gamma(1.6) = 0.89

 

[haruspex]

5. 1/0.7(Gamma1.7) = 1.30 ? Isn't it?

Yes.

I'm sorry questions mess up.It's solution of question 11.

But you haven't done the substitution correctly. What will dx become?

Here is for (8)
For n= 2/3 Gamma(n+1) = Gamma(1+2/3)=Gamma(1.6) = 0.89

OK, but 1.6 is a bit inaccurate. The answer should be > 0.9.

 

[Erbil]

Yes.There's a problem.But I can't figure. if we say u=x^2 --> 2xdx=du --> dx = du/2x Where to go from here? I can't remember how can I escape from x variable?

 

[lurflurf]

For the ones of the type

\int_0^\infty t^a b \, e^{-t^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)

this can be derived by the change of variable u=t^b

 

[haruspex]

Yes.There's a problem.But I can't figure. if we say u=x^2 --> 2xdx=du --> dx = du/2x Where to go from here? I can't remember how can I escape from x variable?

Just replace the x using u=x²

 

[Erbil]

Just replace the x using u=x²

I understand but what about dx ? Would be? x^2=u 2xdx = du dx = du/2x (I'm asking about this x?)

$$ \int _{0}^{\infty }ue^{-u}? $$

 

[lurflurf]

dx = du/(2x)
or
dx = du/(2sqrt(u))

 

[Erbil]

dx = du/(2x)
or
dx = du/(2sqrt(u))

Thanks!

 

[Erbil]

Can I use same logic for 12?

x^3 = u
3x^2dx= du
dx = du / 3*√u

∫xe^-x^3 = ∫u^1/3*e^-u du/3*√u= 1/3 ∫ u^-1/6 e^-udu = 1/3 * Gamma -5/6 ...?+ No idea for 15?

 

[haruspex]

Can I use same logic for 12?

x^3 = u
3x^2dx= du
dx = du / 3*√u

Not √u, a little more complicated than that. What is x as a function of u?

+ No idea for 15?

In line with the other substitutions, u=8x looks obvious. Did you try that?

 

[Erbil]

Not √u, a little more complicated than that. What is x as a function of u?

In line with the other substitutions, u=8x looks obvious. Did you try that?

Oh what I was did.it will be u^1/3 =?
Not yet.But I will.

 

[lurflurf]

^Yes above I said

\int_0^\infty t^a b \, e^{-t^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)

this can be derived by the change of variable u=t^b

also

\int_0^\infty (st)^a b \, e^{-(s t)^b} \, \dfrac{\mathrm{dt}}{t}=\Gamma(a/b)

s>0 this can be derived by the change of variable u=s t