Here is another classic trick, first note that
Now as I stated earlier once
So
So we have that
Now in the first integral we see thta if we let
Then
So we get
So now we see that
Now this is one of the ones that I am most proud of
First let
And we can see that
So we get
Now here is the next step
Let
Now with a little mutliplicative identity
Now imputting the uniformly convergent power seires
We ultimately arrive at (I've doen these earlier in the thread if you are curious)
So now industrious as ever we have found that
Here is another one I have done recently
Now rewrite it as
Now if we let
And
So we have
Now we have done this one before, but all you would do is either use the uniformly convergent Maclaurin series on the denominator and go from there, or use PFD on denominator and do two seperately...either way
No, not as far as proving the series for arcsine. Just an explanation going from there to the zeta solution. I got it to the series you have for arcsin. But I think there are some identities I am not aware of to proceed further. Deriving that zeta(3) out of it is where I am having trouble.
I got this from trying to evaluate ....[1]
Done some fanaggling and found it is equivalent to
I used the series for arcsin and proceeded as follows.
I used the series for arcsin as
I squared that, divided by x and got:
Integrate over 0 to 1 and it becomes:
Now, I was going to proceed with gamma/beta or something, but haven't finished. I see what you mean by tedious. This is where I wasn't totally sure what the best way to go was. I have some ideas, but have not had time to go any further. I was going to look at it after bit. I was asking for your input.
I also found that
is also an equivalent integral.
Playing around with it, I found many approaches.
I was trying to evaluate [1] and started with a u sub of . This lead to
Then, I used the series for and went from there.
gives
Using the aforementioned series and noting that
I got
I then used the gamma/beta thing and whittled it down to
Now using the beta integral , I eventually get:
Then I got hung up. I can find a closed form for the geometric series. That is what I didn't know what to do with. It diverges, so I must've erred somewhere.
Any thoughts there, Big M
Hey MS, you're back. I am sitting here bored. May as well play with this again.
Starting with the series you gave for
We have
Now,
So, we get:
Now, MS, what are your thoughts?. Where is that pesky zeta(3).
I know there is some silly thing I am overlooking.
Here is one that I found, I am sure it has been done before, but I came up with independently and I am crossing my fingers I am lucky enough to notice it first!
I am going to show that
Where is Diriclet's Eta funcion given by
Proof: Consider that
Now note that it is uniformly for every interior point of its interval of convergence. Or seeing that if we define that converges and that . And finally noting that . So we can conclude that this series is uniformly convergent by Abel's Test for Uniform Convergence.
So now we rewrite our integral as follows
Now let us compute
Now letting we get
Now letting gives
So if
We can see by dividing both sides by that
So back to our integral we have
So we can conclude that
Note the restriction on may be extended to fit any points where both and are both well-defined.
Also note that a slight change gives
so many integrals... i think i'm gonna cry.
the first time i saw this integral, i couldn't solve it, but now i came up with a solution and i have to say that this one is a really nice integral.
------
straightforward computations show that and (1)
now so swap the integration order and by applying (1) the integral equals