Evaluate $\int_{0}^{\pi} \frac{x\coth x-1}{x^2}dx$

Solution 1:

An inelegant method (for pure mathematicians) is to use infinite series forms for both sides of the claimed equality to meet at some identical intermediate result:

Starting with the infinte series expansion for $\arctan x$ $$\arctan x = \sum _{k=1}^{\infty } \frac{(-1)^{k-1} }{2 k-1}x^{2 k-1}$$

immediately gives

$$\sum _{r=1}^{\infty } \frac{1 }{r}\arctan\left(\frac{1}{r}\right)=\sum _{r=1}^\infty \frac{1}{r} \sum _{k=1}^{\infty } \frac{(-1)^{k-1} }{2 k-1} \frac{1}{r^{2 k-1}}\tag{1}$$

swapping the order of the double sum gives

$$\sum _{r=1}^{\infty } \frac{1 }{r}\arctan\left(\frac{1}{r}\right)=\sum _{k=1}^{\infty } \frac{(-1)^{k-1} \zeta (2 k)}{2 k-1}\tag{2}$$

Then using the infinite series expansion for $\coth x$ $$\coth(x)=\frac{1}{x}+2 \sum _{k=1}^{\infty } \frac{ (-1)^{k-1} \zeta (2 k)}{\pi ^{2 k}}x^{2 k-1}$$

We can integrate $$\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx$$ term by term to give $$\frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx=\sum _{k=1}^{\infty } \frac{(-1)^{k-1} \zeta (2 k)}{2 k-1}\tag{3}$$

Solution 2:

Approximations.

Since $$\frac{x \coth (x)-1}{x^2}=\sum_{n=0}^\infty \frac{2^{2 n+2} B_{2 n+2} }{(2 n+2)!}x^{2 n}$$ it is simple to build a $[2k,4]$ Padé approximant of it. For example $$P_1= \frac{7 \left(2 x^2+45\right)}{x^4+105 x^2+945}$$ $$P_2=\frac{1}{21}+\frac{43 x^2+990}{7 \left(x^4+60 x^2+495\right)}$$ $$P_3=\frac{7613}{117978}-\frac{x^2}{6678}+\frac{121 \left(166969 x^2+3935295\right)}{39326 \left(106 x^4+5610 x^2+45045\right)}$$ and so on. So, we face the problem of $$I(a,b,c,d)=\int_0^\pi \frac {a x^2+b}{(x^2+c)(x^2+d)}\,dx$$ $$I(a,b,c,d)=\frac 1{c-d} \Bigg[ \frac{ (b-a d)}{\sqrt{d}}\tan ^{-1}\left(\frac{\pi }{\sqrt{d}}\right)-\frac{ (b-a c)}{\sqrt{c}}\tan ^{-1}\left(\frac{\pi }{\sqrt{c}}\right)\Bigg]$$ Now, some numerical results $$\left( \begin{array}{cc} k & \text{approximation} \\ 1 & 0.8949240197 \\ 2 & 0.8950080586 \\ 3 & 0.8950039337 \\ 4 & 0.8950042166 \\ 5 & 0.8950041937 \\ 6 & 0.8950041958 \\ 7 & 0.8950041956 \end{array} \right)$$

Just for the fun of it, an inverse symbolic calculator proposed as an approximation $$I= \Big[e^\alpha \, \pi^\beta \cot ^4(e \pi ) \csc ^{10}(e \pi )\Big]^{\frac 1 {25}}$$ $$\alpha =38+e-\frac{5}{e}-\frac{47}{\pi }-16 \pi\qquad \text{and} \qquad \beta=14 e-19$$ which is in error of $2.29\times 10^{-20}$.

Solution 3:

Another method, if you'd like:

Consider the continued fraction of $\coth x$ which is

$$\coth x=\dfrac{1}{x}\left\{1+\dfrac{x^2}{3+\dfrac{x^2}{5+\dfrac{x^2}{7+\dfrac{x^2}{\ddots}}}}\right\}$$ Considering the third iteration, that is:

$$\dfrac{1}{x}+\dfrac{x}{3+\dfrac{x^2}{5+\dfrac{x^2}{7}}}=\dfrac{1}{x}+\dfrac{5x+\dfrac{x^3}{7}}{15+\dfrac{10x^2}{7}}=\dfrac{105+45x^2+x^4}{5x(21+2x^2)}$$ we may write

$$\int_0^{\pi}\dfrac{105+45x^2+x^4}{5x^2(21+2x^2)}-\dfrac{5(21+2x^2)}{5x^2(21+2x^2)}dx= \int_0^{\pi}\dfrac{35+x^2}{105+10x^2}dx\approx 0.89\color{red}{6287}$$ which was only the third iteration.

Solution 4:

Not an answer, but I thought I'd give some equivalent statements. As James and the original poster mentioned, we have: $$ \sum _{r=1}^{\infty } \frac{1 }{r}\arctan\left(\frac{1}{r}\right)= \sum _{n=0}^{\infty } \frac{(-1)^{n}}{2n+1}\zeta(2n+2) = \frac{\pi}{2}\int_{0}^{\pi} \frac{x\coth x-1}{x^2} \, dx \tag{1} $$

Changing the integrand yields the equivalent: $$ \frac{1}{2}\int_{0}^{1} \frac{\pi\coth (\pi x)}{x} - \frac{1}{x^2} \, dx = \frac{1}{2}\int_{0}^{\infty} \pi\coth (\frac{\pi}{e^x}) - e^x \, dx = (1) $$

Recalling that: $$ \int_{0}^{\infty} e^{-x} \frac{\sin t x}{\sinh x} \, dx = \frac{\pi}{2} \ \coth(\frac{\pi}{2} t) - t $$

and integrating over $t$ with some rearranging gives us $$ \int_{0}^{\infty} \frac{Si(x)}{e^x-1} \, dx = (1) $$ This integral (where Si(x) is the Sine Integral) also equals (1) and part by part can give the series above.

If one expands the $\coth(x)$ in the original integral it can be rearranged to: $$ \frac{\pi}{4} + \frac{1}{4}\int_{0}^{\pi} \psi^{(0)}(-ie^{ix}) \ - \psi^{(0)}(-ie^{-ix}) \ - (-i + 2\cos(x)) \tan(x) \ dx = (1) $$ However, this form is basically useless as the Polygamma function doesn't simplify anything.

I hope others will have better luck than I did.