The Proof is Trivial!

Original post by Smaug123

Is this possible without contour integration? Contour integration is horrible :frown:

Hmm, I'd imagine it would be doable with Taylor's series? Not too sure. :tongue:

Original post by Smaug123

Dare I give:

g(s,t) = 0

,

f(s) = \zeta(s)

?
Apologies if I've misunderstood

\zeta

- I flee from contour integration at every possible opportunity (not that there have been many such opportunities - I really mean that any time it might have been useful to learn contour integration, I ran away from it like the wind).

I've modified the question now - the point of the question being to express the mysterious Zeta function in terms of less-mysterious 'elementary' functions which can, in turn, be used to approximate

\zeta(2n+1)

(

n \in \mathbb{N}

) by using integral approximations :smile:

I learnt it this morning but I'm still feeling a bit shaky (hoping people will engage in discussion on ASOM with me with regards to a few problems!) I'm probably not quite at the point where I could have done the problem LOTF just solved, though I did think it worth posting as I knew a solution would not take long to emerge.

Reply 1922

Hasufel

16

Problem 270

what is:

\displaystyle\int \frac{1}{x^{4}(1+x^{2})^{1/2}}dx

?

Reply 1923

Original post by Hasufel

Problem 270

what is:

\displaystyle\int \frac{1}{x^{4}(1+x^{2})^{1/2}}dx

?

Solution 270

\displaystyle \int \frac{dx}{x^{4} (1+x^2)^{\frac{1}{2}}} \overset{x = \tan t}= \int \cot^{3} t \csc t \ dt

Now, note that

\cot^{2} t + 1 \equiv \csc^{2} t

\displaystyle \begin{aligned} \int \cot^{3} t \csc t \ dt = \int \cot t \csc t ( \csc^2 t - 1) \ dt &\overset{u = \csc t}= \int 1 - u^{2} \ du \\ & = u - \frac{1}{3} u^{3} + \mathcal{C} \\ & = \csc t - \frac{1}{3} \csc^{3} t + \mathcal{C} \\ & = \frac{\sqrt{x^{2} + 1}}{x} - \frac{1}{3} \frac{(x^2 + 1)^{\frac{3}{2}}}{x^3} + \mathcal{C} \\ & = \frac{\sqrt{x^2 + 1} (2x^2 - 1)}{3x^2} + \mathcal{C} \end{aligned}

(edited 10 years ago)

Reply 1924

Original post by Felix Felicis

x

Basically the same as yours but more direct:

Solution 270 (2)

Let

x=\frac{1}{\sqrt{u^2-1}}

,

\displaystyle \Rightarrow \int \frac{1}{x^4 \sqrt{1+x^2}} \ dx = \int (u^2-1)^2 \frac{\sqrt{u^2-1}}{u} \frac{-u}{(u^2-1)^{\frac{3}{2}}} \ du = \int 1- u^2 \ du [br]\displaystyle = u - \frac{1}{3} u^3 + C=\frac{\sqrt{x^2+1} (2x^2-1)}{3x^2} + C

Spoiler

We can generalise by using the same substitution,

\displaystyle \Rightarrow \int \frac{1}{x^{2(n+1)} \sqrt{1+x^2}} \ dx = - \int (u^2-1)^n \ du = - \sum_{r=0}^n \binom{n}{r} (-1)^{n-r} \int u^{2r} \ du [br]\displaystyle \begin{aligned} = - \sum_{r=0}^n \binom{n}{r} (-1)^{n-r} \frac{u^{2r+1}}{2r+1} + C = - \sum_{r=0}^n \binom{n}{r} (-1)^{n-r} \frac{(1+x^2)^{ \frac{2r+1}{2} }}{x^{2r+1}(2r+1)} + C \end{aligned}

(edited 10 years ago)

Reply 1925

\displaystyle \gamma := \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k}-\log(n) \right)

which is known as the Euler-Mascheroni constant.

Problem 271***

Prove that

\displaystyle \int_0^{\infty} e^{-x} \log(x) \ dx = \Gamma'(1) =\psi(1) = -\gamma

.

Note that I require you to prove any non-* theorems/results that you use in your proof.

Problem 272*

Prove that

\displaystyle \begin{aligned} \gamma = -4 \int_0^{\infty} e^{-x^2} x \log(x) \ dx = -\int_0^1 \log \log \left(\frac{1}{x} \right) \ dx = \int_0^{\infty} \left( \frac{1}{e^x-1}-\frac{1}{xe^x} \right) \ dx \end{aligned}[br]\displaystyle = \int_0^1 \left( \frac{1}{\log(x)}+\frac{1}{1-x} \right) \ dx = \int_0^{\infty} \frac{1}{x} \left( \frac{1}{1+x^k} - e^{-x} \right) \ dx

Problem 273***

Evaluate

\displaystyle \int_0^1 \int_0^1 \frac{x-1}{(1-xy) \log(xy)} \ dx

Prove all non-* theorems, as before.

Problem 274***

Evaluate

\displaystyle \int_0^{\infty} e^{-x^2} \log(x) \ dx

Problem 275***

Evaluate

\displaystyle \int_0^{\infty} \frac{1}{x^2} \left( \frac{\log(1+x)}{(\log(x))^2+\pi^2} \right) \ dx

(edited 10 years ago)

Reply 1926

Original post by Mladenov

Problem 165***

Evaluate

\displaystyle \int_{0}^{2} \frac{x^{4}}{(x^{2}+1)(x(2-x)^{3})^{\frac{1}{4}}}dx

.

Not sure this is right as my final answer looks messy, so let me know.

Solution 165

Let

x \mapsto 2x

:

\displaystyle \int_0^2 \frac{(x^2+1)(x^2-1)+1}{(x^{2}+1)(x(2-x)^{3})^{\frac{1}{4}}} \ dx = \int_0^2 \left( \frac{x^2-1}{(x(2-x)^3)^{\frac{1}{4}}}+\frac{1}{(x^2+1)(x(2-x)^3)^{\frac{1}{4}}} \right) \ dx [br]\displaystyle \underbrace{= \int_0^1 \frac{4x^2-1}{x^{\frac{1}{4}}(1-x)^{\frac{3}{4}}} \ dx}_{\alpha} + \underbrace{\int_0^1 \frac{1}{(4x^2+1)(x(1-x)^3)^{\frac{1}{4}}} \ dx}_{\beta}

Comparing the integrals to the Beta function, using the Gamma function representation of the Beta function as well as Euler's reflection formula, we get:

\displaystyle \alpha = 4 \int_0^1 x^{\frac{7}{4}}(1-x)^{-\frac{3}{4}} \ dx - \int_0^1 x^{-\frac{1}{4}} (1-x)^{-\frac{3}{4}} \ dx = 4 B \left(\frac{11}{4},\frac{1}{4} \right) - B \left(\frac{3}{4},\frac{1}{4} \right) [br]\displaystyle = \frac{13}{8} \Gamma \left(1-\frac{1}{4} \right) \Gamma \left(\frac{1}{4} \right)=\frac{13 \pi}{8 \sin(\frac{\pi}{4})}=\frac{13 \pi \sqrt{2}}{8}

Let

x \mapsto \frac{1}{x+1}

and then

x \mapsto x^4

:

\displaystyle \beta = \int_0^{\infty} \frac{1}{(x+1)^2} \left( \frac{1}{(\frac{4}{(x+1)^2}+1)( \frac{x^3}{(x+1)^4})^{\frac{1}{4}}} \right) \ dx = \int_0^{\infty} \frac{x+1}{x^{\frac{3}{4}}(x^2+2x+5)} \ dx [br]\displaystyle = 4 \int_0^{\infty} \frac{x^4+1}{x^8+2x^4+5} \ dx

Splitting the denominator into its complex routes and integrating:

\displaystyle \begin{aligned} = \frac{1}{2} \left[\sum_{ \{w : w^8+2w^4+5=0 \} } \frac{\log(x-w)}{w^3} \right]_0^{\infty} = -\frac{1}{2} \sum_{ \{w: w^8+2w^4+6=0 \} } \frac{\log(-w)}{w^3}=-\frac{\pi}{2} \left( \frac{1+i}{(-1-2i)^{\frac{3}{4}}} + \frac{1-i}{(-1+2i)^{\frac{3}{4}}} \right) \end{aligned}

Simplifying the result as much as we can and combining

\alpha

and

\beta

, we get:

\displaystyle \int_0^2 \frac{x^4}{(x^2+1)(x(2-x)^{3})^{\frac{1}{4}}} \ dx = \pi \left(\frac{13 \sqrt{2}}{8}+5^{-\frac{3}{4}} \left( \frac{2}{-11+15 \sqrt{5}-2 \sqrt{10(25-11 \sqrt{5})}} \right)^{-\frac{1}{4}} \right) \square

Reply 1927

Original post by Jkn

\displaystyle \gamma := \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k}-\log(n) \right)

which is known as the Euler-Mascheroni constant.

Problem 271***

Prove that

\displaystyle \int_0^{\infty} e^{-x} \log(x) \ dx = \Gamma'(1) =\psi(1) = -\gamma

.

Note that I require you to prove any non-* theorems/results that you use in your proof.

Does that include proof of differentiation under the integral sign?

Reply 1928

Original post by Felix Felicis

Does that include proof of differentiation under the integral sign?

Sorry, no! I suppose I meant non-** theorems. Essentially I am asserting that

\psi(1)=-\gamma

is not quotable :smile:

Reply 1929

Sort of a mixture between the last two solutions but I thought the route it leads down seems different enough to warrant a post:

Solution 168 (3)

Beginning as in solution (1), we can deduce that

\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \ln \sin x \ln \cos x \ dx = \frac{\pi}{2} \ln^2 2 - \frac{1}{6} \int_0^{\frac{\pi}{2}} \ln^2 \tan x \ dx \mathop =^{\tan x \mapsto x} \frac{\pi}{2} \ln^2 2- \frac{1}{6} \int_0^{\infty} \frac{\ln^2 x}{1+x^2} \ dx \end{aligned}

Let

\displaystyle \begin{aligned} I(\alpha) = \int_0^{\infty} \frac{x^{\alpha}}{1+x^2} \ dx \mathop =^{x^2 \mapsto x} \frac{1}{2} \int_0^{\infty} \frac{x^{\frac{\alpha-1}{2}}}{1+x} \ dx = \frac{1}{2} B \left(\frac{\alpha+1}{2}, \frac{1-\alpha}{2} \right) = \frac{1}{2} \Gamma \left( \frac{\alpha+1}{2} \right) \Gamma \left( 1-\frac{\alpha+1}{2} \right) \end{aligned}

\displaystyle= \frac{\pi}{2} \csc \frac{\pi}{2} (\alpha+1)

\displaystyle \begin{aligned} \Rightarrow \int_0^{\infty} \frac{x^{\alpha} \ln^2 x}{1+x^2} \ dx = \frac{\pi}{2} \frac{d^2}{d \alpha^2} \csc \frac{\pi}{2} (\alpha+1) = \frac{\pi^3}{8} \left(\csc \frac{\pi}{2} (\alpha+1) \cot^2 \frac{\pi}{2} (\alpha+1)+\csc^3 \frac{\pi}{2} (\alpha+1) \right) \end{aligned}

\displaystyle \Rightarrow \int_0^{\frac{\pi}{2}} \ln \sin x \ln \cos x \ dx = \frac{\pi}{2} \ln^2 2 - \frac{1}{6} I''(0) = \frac{\pi}{2} \ln^2 2 - \frac{\pi^3}{48} \ \square

We can generalise the last integral in a rather interesting way:

\displaystyle \begin{aligned} \int_0^{\frac{\pi}{2}} \ln^n \tan x \ dx = \int_0^{\infty} \frac{\ln^n x}{1+x^2} \ dx = \frac{\pi}{2} \frac{d^n}{d \alpha^n} \csc \frac{\pi}{2} (\alpha+1) \Bigg|_{\alpha=0} = \frac{\pi^{n+1}}{2^{n+1}} \frac{d^n}{d \alpha^n} \csc \alpha \Bigg|_{\alpha=\frac{\pi}{2}} \end{aligned}

We now note the Taylor's series of

\csc x

about

x=\frac{\pi}{2}

\displaystyle \csc x = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!}E_{2k} \left(z-\frac{\pi}{2} \right)^{2k}

for

|z-\frac{\pi}{2}|<\frac{\pi}{2}

where

E_n

denotes the nth Euler number.

By considering the coefficients of this series, we can find closed forms for the above integral:

\displaystyle \int_0^{\infty} \frac{\ln^n x}{1+x^2} \ dx = \frac{(-1)^{\frac{n}{2}} E_{n} \pi^{n+1}}{2^{n+1}}

for all

n \in \mathbb{N}^{*}

.

Note that this implies that the above integral is equal to 0 whenever n is odd (this is implied both by the Taylor's series and the fact that all nth Euler numbers are 0 when n is odd).

So, for example,

\displaystyle \int_0^{\infty} \frac{\ln^8 x}{1+x^2} \ dx = \frac{1385 \pi^9}{2^9}

Corollary: Letting n=0 gives us yet another way to deduce that

\displaystyle \int_0^{\infty} \frac{1}{1+x^2} \ dx = \frac{E_0 \pi}{2}=\frac{\pi}{2}

(edited 10 years ago)

Reply 1930

Solution 274

Let

I = \displaystyle \int_{0}^{\infty} e^{-x^2} \ln x \ dx \overset{x \mapsto \sqrt{x}}= \frac{1}{4} \int_{0}^{\infty} x^{-\frac{1}{2}} e^{-x} \ln x \ dx

Now, consider the gamma function:

\displaystyle \begin{aligned} \Gamma (\lambda) = \int_{0}^{\infty} x^{\lambda - 1} e^{-x} \ dx \implies \Gamma ' (\lambda ) & = \int_{0}^{\infty} \frac{\partial}{\partial \lambda} x^{\lambda - 1} e^{-x} \ dx \\ & = \int_{0}^{\infty} x^{\lambda - 1} e^{-x} \ln x \ dx \end{aligned}

which evaluated at

\lambda = \frac{1}{2}

gives

4I

We have that

\displaystyle \begin{aligned} \psi (\lambda ) = \dfrac{\Gamma ' (\lambda)}{\Gamma ( \lambda )} \implies \Gamma ' (\lambda ) & = \psi (\lambda ) \Gamma (\lambda ) \\ \implies \Gamma ' \left( \dfrac{1}{2} \right) & = \Gamma \left( \dfrac{1}{2} \right) \psi \left( \frac{1}{2} \right) \\ & = - \sqrt{\pi} \left( \gamma + 2 \ln 2 \right) = 4 I \end{aligned}

\displaystyle \therefore \int_{0}^{\infty} e^{-x^2} \ln x \ dx = - \frac{1}{4} \sqrt{\pi} \left( \gamma + 2 \ln 2 \right)

(edited 10 years ago)

Reply 1931

Original post by Felix Felicis

x

Nice stuff man, don't you just love gamma functions, DUTIS and all that crap? :biggrin:

Btw, do you like my two monster posts? :colone:

Reply 1932

Original post by Jkn

Nice stuff man, don't you just love gamma functions, DUTIS and all that crap? :biggrin:

Btw, do you like my two monster posts? :colone:

Gotta love it man, it's some good stuff :sogood:

I do! Very nice :colone:

I especially liked the extension to 168 :biggrin:

And while I'm at it, I was looking through some old stuff and saw your method on analysing the Gaussian integral, that was quite nice too :biggrin:

Reply 1933