Finding the value of $\beta(5)$ via a definite integral

Question

I was trying to compute the integral:

$$ I = \int_0^1 \frac{(\operatorname{ln}x)^4}{1+x^2} dx$$

This is not a very difficult integral to evaluate if one knows the standard procedure used in evaluating such types of integral.

I substituted $-\operatorname{ln}x = t$ and the integral becomes:

$$ I = \int_0^{\infty} \frac{t^4 \cdot e^{-t}}{1+e^{-2t}} dt$$

$$ I = \int_0^{\infty} t^4 \cdot e^{-t} \cdot \sum_{n=0}^{\infty} (-1)^n \cdot e^{-2nt} dt$$

$$ I = \sum_{n=0}^{\infty} (-1)^n \cdot \int_0^{\infty} t^4 \cdot e^{-(2n+1)t} dt$$

$$ I = \sum_{n=0}^{\infty} (-1)^n \cdot \frac{(4!)}{(2n+1)^5} = 24\beta(5)$$

When I checked the value of $I$ in WolframAlpha, it suggested that it is equal to $\frac{5 \pi^5}{64}$.

So that must mean

$$ \beta(5) = \frac{5 \pi^5}{24 \cdot 64} = \frac{5 \pi^5}{1536}$$

Which is indeed true.

What I want to know is - Is there any other way to compute $\beta(5)$, like how we compute $\zeta(2)$?

Note that the $\beta$ function in this question refers to the Dirichlet beta function, not to be confused with the Eulerian Beta function.

Answer 1

Polygamma Function

The polygamma function of order $(m-1)$ is defined as the $m$-th derivative of $(\ln \Gamma(x))$:

$$ \psi^{(m-1)}(x) = \frac{d^m}{dx^m} \ln \Gamma(x) $$

Reflection Relation

The reflection relation between polygamma functions is given by

$$ (-1)^m \psi^{(m)}(1-z) - \psi^{(m)}(z) = \pi \frac{\mathrm{d}^m}{\mathrm{d} z^m} \cot(\pi z). $$

It can be easily derived by differentiating the reflection relation of the gamma function $m$ times:

$$ \frac{d^m}{dz^m} \ln\left(\Gamma(z) \Gamma(1-z)\right) = \frac{d^m}{dz^m} \ln\left(\frac{\pi}{\sin(\pi z)}\right) $$

Dirichlet Beta Function

The Dirichlet beta function is defined as:

$$ \beta(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s} $$

If we separate terms like this:

$$ \beta(s) = \sum_{n=0}^\infty \frac{1}{(4n + 1)^s} - \sum_{n=0}^\infty \frac{1}{(4n + 3)^s} $$

Then, it can be expressed in terms of the Hurwitz zeta function and polygamma function as well:

$$ \beta(s) = 4^{-s} \left( \zeta\left(s,\frac{1}{4}\right)-\zeta\left(s,\frac{3}{4}\right) \right) $$

Polygamma series expansion

For the polygamma function, let us recall(derive) the series expansion:

$$ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \ln \Gamma(x) $$ diffrentiating

$$ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \sum_{k=1}^{x} \frac{1}{k} $$ adding and subtracting $ \sum_{k=x+1}^{\infty} \frac{1}{k} $ of eqn

$$ \psi^{(m)}(x) = \frac{d^{(m+1)}}{dx^{(m+1)}} \left(\sum_{k=1}^{x} \frac{1}{k}+ \sum_{k=x+1}^{\infty} \frac{1}{k} - \sum_{k=x+1}^{\infty} \frac{1}{k}\right) $$ now $$ \psi^{(m)}(x) =\frac{d^{(m+1)}}{dx^{(m+1)}} \left(\sum_{k=1}^{\infty} \frac{1}{k} + \sum_{k=1}^{\infty} \frac{1}{x+k}\right) $$

now we can see that first term inside brackets will be treated as constant because its converging and we can easily prove it by integral test or also approximate for more info see Harmonic Series and by differentiating it m times we get below expression $$ \psi^{(m)}(z) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{(z+k)^{m+1}} $$

Now, considering specific values for $z = 1/4$ and $z = 3/4$:

$$ \psi^{(m)}\left(\frac{1}{4}\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{\left(\frac{1}{4}+k\right)^{m+1}} $$

$$ \psi^{(m)}\left(\frac{3}{4}\right) = (-1)^{m+1}\, m! \sum_{k=0}^\infty \frac{1}{\left(\frac{3}{4}+k\right)^{m+1}} $$

Then, subtracting these two equations gives us the relation:

$$ \beta(s) = \frac{1}{2^s} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{s}} = \frac1{(-4)^s(s-1)!}\left[\psi^{(s-1)}\left(\frac{1}{4}\right)-\psi^{(s-1)}\left(\frac{3}{4}\right)\right] $$

If (s) is odd, then:

$$ \beta(2m+1) = \frac{1}{2^{2m+1}} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{2m+1}} = \frac1{(-4)^{2m+1}(2m)!}\left[\psi^{(2m)}\left(\frac{1}{4}\right)-\psi^{(2m)}\left(\frac{3}{4}\right)\right] $$

By using the reflection formula of the polygamma function, we obtain the value of:

$$ (-1)^m \psi^{(m)}\left(1-\frac{1}{4}\right) - \psi^{(m)}\left(\frac{1}{4}\right) = \left. \frac{\pi \, \mathrm{d}^{(m)}}{\mathrm{d} z^{(m)}} \cot(\pi z) \right|_{z=\frac{1}{4}} $$

Which gives us the formula:

$$ \beta(2m+1) = \frac{1}{2^{2m+1}} \sum_{n=0}^\infty\frac{(-1)^{n}}{\left(n+\frac{1}{2}\right)^{2m+1}} = \frac{1}{(4)^{2m+1}(2m)!}\Bigg|_{z=\frac{1}{4}} \frac{\pi \, \mathrm{d}^{(2s)}}{\mathrm{d} z^{(2s)}} \cot(\pi z) $$

Then:

$$ \beta(5) = \frac{5\pi^5}{2(2^5)(4)!} $$

Booooooooooom!

Answer 2

We prove this by computing the integral in anther way. First observe that by substituting $x=\frac 1u$, we get $$I = \int_1^\infty \frac{\ln(u)^4}{1+u^2} du$$ Therefore $$I = \frac{1}{2}\int_0^\infty \frac{\ln(x)^4}{1+x^2} du$$

We can solve this with contour integration. Consider the integral of $f(z)=\frac{A\log(z)^5+B\log(z)^3+C\log(z)}{z^2+1}$ over the keyhole contour. Here $\log$ is taken to have a branch cut on the negative real line.

i

Using standard $ML$ lemma arguments, one can show that the circular arcs of the contour vanish in the $R\to \infty$ and $r\to 0$ limits. Therefore, the only two contributions come from the $M$ and $N$ contours.

We can directly parameterize the $M$ and $N$ contour with $z = -t+i\epsilon$ and $z=-t-i\epsilon$ for $0<t<\infty$, respectively. As $\epsilon$ approaches zero, notice that the branch cut causes $\log(z)$ to approach different values. It approaches $ln(t)+i\pi$ on $M$ and $\ln(t)-i\pi$ on $N$.

The contour integral becomes $$\oint_\gamma f(z)dz = \oint_M f(z)dz+\oint_N f(z)dz $$$$=\int_0^\infty \frac{A(\ln(t)+i\pi)^5+B(\ln(t)+i\pi)^3+C(\ln(t)+i\pi)}{1+t^2}dt -\int_0^\infty \frac{A(\ln(t)-i\pi)^5+B(\ln(t)-i\pi)^3+C(\ln(t)-i\pi)}{1+t^2}dt$$

Now, the odd powers of $\ln(t)$ cancel and we are left with only the even powers. $$=\int_0^\infty \frac{ 10 π A \ln(t)^4i+ 2 \pi (3 B - 10 \pi^2 A) \ln(t)^2i+ 2 \pi (\pi^4 A - \pi^2 B + C)i}{1+t^2}dt $$

By matching the coefficients in the powers of $\ln(t)$, we now can solve a linear system to find the values of $A,B,C$ such that this contour integral is equal to $I$ and get $A = -\frac{1}{20\pi}i$, $B=-\frac{\pi}6i$ and $C=-\frac{7}{60}\pi^3i$.

On the other hand, we can use the residue theorem to compute to contour integral. We have two simple poles at $\pm i$ with the following residues:

$$\mathrm{Res}(f,i)=\frac{A\log(i)^5+B\log(i)^3+C\log(i)}{2i}=-\frac{5\pi^4}{256}i$$

$$\mathrm{Res}(f,-i)=\frac{A\log(-i)^5+B\log(-i)^3+C\log(-i)}{-2i}=-\frac{5\pi^4}{256}i$$

We can now compute $I$ as

$$I = \int_\gamma f(z)dz= 2\pi i (\mathrm{Res}(f,i)+\mathrm{Res}(f,-i))$$ $$ \implies I = 4\pi i (-\frac{5\pi^4}{256}i) = \frac{5\pi^5}{64} $$

Q.E.D.

Answer 3

If you are comfortable with polylogarithms, the antiderivative is not too bad (have a look here).

Using the bounds, the only thing which remains is $$12 i (\text{Li}_5(-i)-\text{Li}_5(i))$$ which is the number.

Edit

If you make it more general $$I_n=\int_0^1\frac{\log ^n(x)}{x^2+1}\,dx$$ looking at the patterns (tedious work) $$I_n=(-1)^n \,\frac {n!}{4^{n+1} }\,\left(\zeta \left(n+1,\frac{1}{4}\right)-\zeta \left(n+1,\frac{3}{4}\right)\right)$$ where appears the generalized Riemann zeta function.

Answer 4

For any positive odd integer $s$, we can express $\beta(s)$ in terms of the derivative of a cotangent function. $$ \begin{aligned} \beta(s)&=\sum_{n=0}^{\infty} \frac{(-1)^n}{(2 n+1)^s} \\ & =\sum_{n=0}^{\infty}\left[\frac{1}{(4 n+1)^s}-\frac{1}{(4 n+3)^s}\right] \\ & =\sum_{n=0}^{\infty}\left[\frac{1}{(4 n+1)^s}+\frac{1}{(-4 n-3)^s}\right] \\ & =\sum_{n=0}^{\infty}\left[\frac{1}{(4 n+1)^s}+\frac{1}{(4(-n-1)+1)^s}\right] \\ & =\sum_{n \in \mathbb{Z}} \frac{1}{(4 n+1)^s} \\ & =\frac{1}{4^s} \sum_{n \in \mathbb{Z}} \frac{1}{\left(n+\frac{1}{4}\right)^s} \end{aligned} $$ for any $x\not \in \mathbb{Z}$, differentiating the identity w.r.t. $x$ $$\pi \cot (\pi x)=\sum_{n \in \mathbb{Z}} \frac{1}{n+x} $$ by $s-1$ times yields $$ \boxed{\beta(s)=\left.\frac{\pi^s(-1)^{s-1}}{(s-1) ! 4^s} \frac{d^{s-1}}{d x^{s-1}}[\cot (x)]\right|_{x=\frac{\pi}{4}}} $$ In particular, when $s=5$, we have $$ \beta(5)=\left.\frac{\pi^5}{4 ! 4^5} \frac{d^4}{dx^4}(\cot x)\right|_{x=\frac{\pi}{4}}=\frac{80 \pi^5}{24 \cdot 4^5}=\frac{5 \pi^5}{1536} $$

Answer 5

$n\geq 1$,integer, \begin{align} B(n)&=\int_0^1\frac{\ln^n x}{1+x^2}dx\\ \int_0^\infty\frac{\ln^{2n}x}{1+x^2}dx&=\int_0^1\frac{\ln^{2n}x}{1+x^2}dx+\underbrace{\int_1^\infty\frac{\ln^{2n}x}{1+x^2}dx}_{u=\frac{1}{x}}=2\text{B}(2n) \end{align}

\begin{align} A&=\int_0^\infty\int_0^\infty\frac{\ln^4(xy)}{(1+x^2)(1+y^2)}dxdy\\ &\overset{u(x)=xy}=\int_0^\infty\int_0^\infty\frac{y\ln^4 u}{(u^2+y^2)(1+y^2)}dudy\\ &=\frac{1}{2}\int_0^\infty \left[\frac{\ln\left(\frac{u^2+y^2}{1+y^2}\right)}{1-u^2}\right]_{y=0}^{y=\infty}\ln^4udu\\ &=-\int_0^\infty\frac{\ln^5 u}{1-u^2}du=-\int_0^1\frac{\ln^5 u}{1-u^2}du-\underbrace{\int_1^\infty\frac{\ln^5 u}{1-u^2}du}_{z=\frac{1}{u}}\\ &=-2\int_0^\infty\frac{\ln^5 u}{1-u^2}du=\boxed{\frac{945}{4}\zeta(6)} \end{align} On the other hand, \begin{align}A&=\int_0^\infty\int_0^\infty\frac{\ln^4 y+4\ln x\ln^3y+6\ln^2x\ln^2y+4\ln^3x\ln y+\ln^4 x}{(1+x^2)(1+y^2)}dxdy\\ &=\boxed{2\pi\text{B}(4)+24\text{B}(2)^2} \end{align} Therefore, \begin{align}\boxed{\text{B}(4)=\frac{945\zeta(6)-96\text{B}(2)^2}{8\pi}}\end{align} Since, \begin{align}\text{B}(2)&=\frac{\pi^3}{16},\zeta(6)=\frac{\pi^6}{945}\end{align} then, \begin{align}\boxed{\text{B}(4)=\frac{5}{64}\pi^5}\end{align}

NB: \begin{align}\int_0^\infty \frac{\ln x}{1+x^2}dx=0\end{align}