Recent content by SumThePrimes

It is (e^x)+1 . I tried using limits on the bounds and then integration by parts and then switching certain values to other limited integrals and canceling... I tell you what that didn't work. I also tried a limit for the natural logarithm with an integral for the Dirichlet eta function, but...

Hello, I have recently encountered an integral that I have been able to evaluate in a sick, unholy way, and for making a proof much more elegant I would like a simple way to evaluate the integral from 0 to infinity of ln(x)/(e^x+1) . thank you!

If f(x)=1+1x+2x^2+3x^3+5x^4+8x^5... f(x)*x=1x+1x^2+2x^3+3x^4+5x^5... f(x)*x^2 =1x^2+1x^3+2x^4+3x^5... Then f(x)-x*f(x)-x^2*f(x)=1 , so f(x)(1-x-x^2)=1 , so f(x)=1/(1-x-x^2) Your series is this function evaluated at x=1/2 , which has a value of 4. Isn't that something? :)

I've recently been confronted with the limit as n goes to 0 of sin(\pi*n/4)*\Gamma(x) , and have no idea on how to confront the problem, as I have little familiarity with the gamma function. Is there any relatively easy ways to prove this, or at least ways that use methods not difficult to...

For one that satisfies only the second statement, if f(a+b)=f(a*b), then f(a+0)=f(a*0), so f(a) is f(0) for any a, so the second statement alone assures us that this function is constant, but there is no way to know what constant with only the fact f(a+b)=f(a*b) . Now combine your second...

When I said your number I was referring to the original posts number, not necessarily your clarification of it. I thought I knew the syntax with these towers, I know tetration starts at the top, it probably is best to use grouping symbols, as you say. Although I am rather sure in general you...

I always though with exponent towers you start at the top, not the bottom, I'm rather sure that's how you work with those... Regardless, your number is less than Grahams number.

Do you mean is (2^3^4^5^6^7^8^9)^(2^3^4^5^6^7^8^9) greater than grahams number? I don't have a proof, but I know what Grahams number is and your number is so much smaller that there is no word to compare them... Grahams number is so huge, it's crazy.

Wow, thank you, I am acquainted with the sines of imaginary numbers and that formula, but it never even crossed my mind... So much for e^(4*pi^2)-1 ... I wouldn't have thought an exact solution... Maybe Euler is better at infinite products than me(Just maybe...:rolleyes: ), I actually got this...

I was wondering about the product n=1 to infinity of (1+n^-2) , I used a very unorthodox, to say the least, manipulation of complex numbers that shows that it should equal a particular number larger than 10^17 , however wolfram alpha can't seem to give me a answer, and when I sum to very large...

Ah, thank you very much, I should have consulted wolfram alpha. I'm sure I'll have enough for a proof of it now, thank you! My summing of divergent resulted in the common answer 1+2+3+4...=-1/12, though I can now compute 2*1-4*3+6*5-8*7..., or indeed any n!/1-(n+3)!/3!+(n+5)!/5!-(x+7)!/7...

Well I have tried integration by parts on it, but it seemed to just explode to astronomical sizes... I have calculated it for certain values of n, and am rather sure that it can be expressed with common functions for any integer, and in particular I only need the Integral from 0 to infinity of...

I have recently come up with a fascinating method of summing various divergent series, whether it has been done before I am unaware, but in some scenarios it involves the anti-derivative of e^(-x)*sin(x)*x^n with respect to x. I only need positive integer values of n, and of course a...