# Thread: Prove some identities! (Round two)

1. Originally Posted by Drexel28
So then I suppose it will be impressive when you prove it, right!
Spoiler:

Define $\displaystyle f(z) = \frac d{dz}\left(\log\Gamma(z)\right) = \frac{\Gamma'(z)}{\Gamma(z)}$

From the Gamma function duplication formula we see $\displaystyle f(z)+f(z+\tfrac12)+2\log2 = 2f(2z)$. Thus $\displaystyle f(\tfrac32) = 2f(2)-f(1)-2\log2$.

Now we must calculate $\Gamma'(1)$ and $\Gamma'(2)$. Using a definition of $\Gamma$, we see $\displaystyle f(s) = -\frac1s-\gamma-\sum_{n=1}^\infty\left(\frac1{n+s}-\frac1n\right)$

From this it's easy to see $\displaystyle f(1) = -\gamma$ and $\displaystyle f(2) = 1-\gamma \quad\text{\footnotesize(telescoping sum)}$.

So we see $f(\tfrac32) = 2-\gamma-2\log2$

-----

We know $\displaystyle \Gamma(z+1) = z\Gamma(z) \implies f(x+1) + f(x)+\frac1x$. Thus $f(\tfrac12) = f(\tfrac32)-2 = -\gamma-2\log2$.

Finally $\displaystyle \Gamma'\left(\frac12\right) = \Gamma\left(\frac12\right)(-\gamma-2\log2) = -\sqrt{\pi}(\gamma+2\log2)$.

2. Originally Posted by chiph588@
Spoiler:

Define $\displaystyle f(z) = \frac d{dz}\left(\log\Gamma(z)\right) = \frac{\Gamma'(z)}{\Gamma(z)}$

From the Gamma function duplication formula we see $\displaystyle f(z)+f(z+\tfrac12)+2\log2 = 2f(2z)$. Thus $\displaystyle f(\tfrac32) = 2f(2)-f(1)-2\log2$.

Now we must calculate $\Gamma'(1)$ and $\Gamma'(2)$. Using a definition of $\Gamma$, we see $\displaystyle f(s) = -\frac1s-\gamma-\sum_{n=1}^\infty\left(\frac1{n+s}-\frac1n\right)$

From this it's easy to see $\displaystyle f(1) = -\gamma$ and $\displaystyle f(2) = 1-\gamma \quad\text{\footnotesize(telescoping sum)}$.

So we see $f(\tfrac32) = 2-\gamma-2\log2$

-----

We know $\displaystyle \Gamma(z+1) = z\Gamma(z) \implies f(x+1) + f(x)+\frac1x$. Thus $f(\tfrac12) = f(\tfrac32)-2 = -\gamma-2\log2$.

Finally $\displaystyle \Gamma'\left(\frac12\right) = \Gamma\left(\frac12\right)(-\gamma-2\log2) = -\sqrt{\pi}(\gamma+2\log2)$.

3. I don't understand how you used the duplication formula. Could you explain? I'm stupid.

Originally Posted by chiph588@
Spoiler:

Define $\displaystyle f(z) = \frac d{dz}\left(\log\Gamma(z)\right) = \frac{\Gamma'(z)}{\Gamma(z)}$

From the Gamma function duplication formula we see $\displaystyle f(z)+f(z+\tfrac12)+2\log2 = 2f(2z)$. Thus $\displaystyle f(\tfrac32) = 2f(2)-f(1)-2\log2$.

Now we must calculate $\Gamma'(1)$ and $\Gamma'(2)$. Using a definition of $\Gamma$, we see $\displaystyle f(s) = -\frac1s-\gamma-\sum_{n=1}^\infty\left(\frac1{n+s}-\frac1n\right)$

From this it's easy to see $\displaystyle f(1) = -\gamma$ and $\displaystyle f(2) = 1-\gamma \quad\text{\footnotesize(telescoping sum)}$.

So we see $f(\tfrac32) = 2-\gamma-2\log2$

-----

We know $\displaystyle \Gamma(z+1) = z\Gamma(z) \implies f(x+1) + f(x)+\frac1x$. Thus $f(\tfrac12) = f(\tfrac32)-2 = -\gamma-2\log2$.

Finally $\displaystyle \Gamma'\left(\frac12\right) = \Gamma\left(\frac12\right)(-\gamma-2\log2) = -\sqrt{\pi}(\gamma+2\log2)$.

4. Originally Posted by Random Variable
I don't understand how you used the duplication formula. Could you explain. I'm stupid.
You're not stupid, math is hard!

$\displaystyle 2\sqrt{\pi}\Gamma(2z) = 2^{2z}\Gamma(z)\Gamma(z+\tfrac12) \implies \log(2\sqrt{\pi})+\log\Gamma(2z) = \log 2^{2z} + \log\Gamma(z) + \log\Gamma(z+\tfrac12)$

$\displaystyle \frac d{dz}\left(\log(2\sqrt{\pi})+\log\Gamma(2z)\right) = \frac d{dz}\left(2z\log 2 + \log\Gamma(z) + \log\Gamma(z+\tfrac12)\right)$

$\displaystyle \implies 2f(2z) = 2\log2+f(z)+f(z+\tfrac12)$

5. More "fun" with the polygamma function.

EDIT: Show that $\displaystyle \int^{\infty}_{0} \sin (x) x^{a-1} \ dx = \Gamma(a) \sin \big(\frac{\pi a}{2} \big) \ 0 .

Then use the result to find $\displaystyle \int^{\infty}_{0} \sin(x^{2}) \ln^{2} x \ dx$ .

6. Originally Posted by Random Variable
More "fun" with the polygamma function.

EDIT: Show that $\displaystyle \int^{\infty}_{0} \sin (x) x^{a-1} \ dx = \Gamma(a) \sin \big(\frac{\pi a}{2} \big) \ 0 .

Then use the result to find $\displaystyle \int^{\infty}_{0} \sin(x^{2}) \ln^{2} x \ dx$ .
I'll leave the second part to someone else:

Spoiler:

Let

$\displaystyle I(a)=\int_0^{\infty}\frac{\sin(x)}{x^{1-a}}\text{ }dx$

Note then that

$\displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\sin(x)\int_0^{\infty} y^{-a}e^{-xy}dy dx$

Revering the order of integration gives

$\displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\int_0^{\infty}\sin(x) y^{-a}e^{-xy} dx dy=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\frac{y^{-a}}{y^2+1}\text{ }dy$

This last integral is relatively easy (we've done it in this or the last thread I think) and thus we get that

\displaystyle \begin{aligned}I(a) &=\frac{1}{\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &=\frac{\Gamma(a)}{\Gamma(a)\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \frac{\Gamma(a)}{\frac{\pi}{\sin(\pi a)}}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\frac{\sin(\pi a)}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\sin\left(\frac{\pi a}{2}\right)\end{aligned}

7. I think the integral also converges for $-1. But $a=0$ seems to be a problem even if you take the limit as $a$ goes to zero.

EDIT: I didn't make that clear. I meant before you manipulated your answer, should it be $\cos \Big( \frac{\pi a}{2} \Big)$ or $\sin \Big(\frac{\pi a}{2} \Big)$ ?

Originally Posted by Drexel28
I'll leave the second part to someone else:

Spoiler:

Let

$\displaystyle I(a)=\int_0^{\infty}\frac{\sin(x)}{x^{1-a}}\text{ }dx$

Note then that

$\displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\sin(x)\int_0^{\infty} y^{-a}e^{-xy}dy dx$

Revering the order of integration gives

$\displaystyle I(a)=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\int_0^{\infty}\sin(x) y^{-a}e^{-xy} dx dy=\frac{1}{\Gamma(1-a)}\int_0^{\infty}\frac{y^{-a}}{y^2+1}\text{ }dy$

This last integral is relatively easy (we've done it in this or the last thread I think) and thus we get that

\displaystyle \begin{aligned}I(a) &=\frac{1}{\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &=\frac{\Gamma(a)}{\Gamma(a)\Gamma(1-a)}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \frac{\Gamma(a)}{\frac{\pi}{\sin(\pi a)}}\frac{\pi}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\frac{\sin(\pi a)}{2\cos\left(\frac{\pi a}{2}\right)}\\ &= \Gamma(a)\sin\left(\frac{\pi a}{2}\right)\end{aligned}

8. Originally Posted by Random Variable
I think the integral also converges for $-1. But $a=0$ seems to be a problem even if you take the limit as $a$ goes to zero.
It converges for $\text{Re}(a)\in(-1,1)$. When, $a=0$ this is the famous integral $\displaystyle \int_0^{\infty}\frac{\sin(x)}{x}\text{ }dx=\frac{\pi}{2}$.

And should it be $\cos \Big( \frac{\pi a}{2} \Big)$ or $\sin \Big(\frac{\pi a}{2} \Big)$ ?
Are you talking about for the integral extended to negative values, or for my integral? For mine it is supposed to be $\cos\left(\frac{\pi a}{2}\right)$? Would you like a proof?

9. I'm well aware of that integral. But does $\lim_{a \to 0} \Gamma(a) \sin \Big( \frac{\pi a}{2} \Big) = \frac{\pi}{2}$ ?

And I thought $\displaystyle \int^{\infty}_{0} \frac{y^{-a}}{1+y^{2}} \ dx = \frac{\frac{\pi}{2}}{\sin \frac{\pi (1-a)}{2}} = \frac{\frac{\pi}{2}}{\cos \frac{\pi a}{2}}$. Nevermind.

10. Originally Posted by Random Variable
I'm well aware of that integral. But does $\lim_{a \to 0} \Gamma(a) \sin \Big( \frac{\pi a}{2} \Big) = \frac{\pi}{2}$ ?
Go back to extending $\Gamma$ analytically to $\mathbb{C}$ by the reflection formula! In particular, in my methodology above we see that $\displaystyle \Gamma(a)\sin\left(\frac{\pi a}{2}\right)=\frac{\pi}{2}\frac{\Gamma(1-a)}{\cos\left(\frac{\pi a}{2}\right)}$. What happens if you plug $a=0$ in there?

11. For $\displaystyle k\in\mathbb{Z}$, evaluate $\displaystyle \sum_{n=1}^\infty (-1)^nn^{-1+2\pi ik/\log2}$

Not too difficult, but the result is somewhat surprising if you've never seen it.

12. Originally Posted by chiph588@
For $\displaystyle k\in\mathbb{Z}$, evaluate $\displaystyle \sum_{n=1}^\infty (-1)^nn^{-1+2\pi ik/\log2}$

Not too difficult, but the result is somewhat surprising if you've never seen it.
This was surprising!

Spoiler:

Merely note that for $k\ne 0$ we have that

$\displaystyle \sum_{n=1}^{\infty}(-1)^n n^{-1+\frac{2\pi i k}{\log(2)}}=\left(1-2^{\frac{2\pi i k}{\log(2)}\right)\zeta\left(1-\frac{2\pi i k}{\log(2)}\right)=\left(1-1\right)\zeta\left(1-\frac{2\pi i k}{\log(2)}\right)=0$

and for $k=0$ this is just the alternating harmonic series which evaluates to $\log(2)$. It follows that

$\displaystyle \sum_{n=1}^{\infty}(-1)^{n}n^{-1+\frac{2\pi i k}{\log(2)}}=\begin{cases}0 & \mbox{if}\quad k\ne0\\ \log(2) & \mbox{if}\quad k=0\end{cases}$

13. Try this:

$\displaystyle \sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_nn!}\quad c-b-a>0$

14. Originally Posted by Drexel28
Try this:

$\displaystyle \sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{(c)_nn!}\quad c-b-a>0$
That's just Gauss' hypergeometric series...

This thread has started as "Prove some identities", and has become "Use this transcendental function's properties as proven in Whittaker and Watson". There's nothing wrong with it, but it'd be fun to go back to some relatively simple stuff.

Here's an easy problem I made up : show that, for any positive integer $n$, we have

$\displaystyle \sum_{k=2}^\infty\lfloor \log_k n\rfloor = \sum_{k=1}^\infty \lfloor n^{1/k}-1\rfloor$

15. I didn't mean to kill this thread! Somebody must have solved the problem I gave but forgotten to post their solution... right? ^^

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