A while back I was working through many problems in Mathews and Walker's Mathematical Methods of Physics. In the appendix is this problem:

A-6. Find the residue of the function $z^2 e^{1/\sin z}$ at the isolated (essential) singularity $z=\pi$.

It is not too hard to get this residue numerically (it is $-7.5764\cdots$). I struggled for many pages to find an exact result. Does an analytical expression for this residue exist and, if so, what is it? (And briefly, how did you get it?)

  • 8
    $\begingroup$ Even though the problem comes from a book intended primarily for physicists, I don't think that the [mathematical-physics] tag is relevant here. $\endgroup$ – Adrián Barquero Mar 14 '12 at 23:44

It looks like you're not the first one to despair of this problem – it's been asked in various places on the net and never got an answer.

It always depends on what you call an analytical expression, but using Lagrange inversion you can express the residue in terms of series, some of which can be written as simple integrals over modified Bessel and Struve functions. The basic Lagrange inversion formula to be used is

$$k[z^k](\arcsin z)^n=n[z^{-n}](\sin z)^{-k}\;,$$

where $[z^m]$ denotes extracting the coefficient of $z^m$ in the Laurent series around $z=0$. Using $\sin (\pi+x)=-\sin x$, we can write the desired residue as a residue at $z=0$:

$$ \def\res#1#2{\operatorname{Res}_{z=#1}\left(#2\right)} \res{\pi}{z^2\mathrm e^{1/\sin z}}=\res0{(z+\pi)^2\mathrm e^{-1/\sin z}}\;. $$

To calculate this, we need to calculate the coefficients of $z^{-1}$, $z^{-2}$ and $z^{-3}$ in the Laurent series for $\mathrm e^{-1/\sin x}$. Using the Taylor series for $\arcsin z$,

$$\arcsin z=\sum_{m=0}^\infty\frac{(2m)!}{(m!2^m)^2(2m+1)}z^{2m+1}\;,$$

the coefficient of $z^{-1}$ is obtained as

$$ \def\hypergeom#1#2#3#4#5{{}_{#1}F_{#2}(#3;#4;#5)} \begin{eqnarray} [z^{-1}]\mathrm e^{-1/\sin z} &=& [z^{-1}]\sum_k\frac{(-\sin z)^{-k}}{k!} \\ &=&\sum_k(-1)^k\frac{[z^{-1}](\sin z)^{-k}}{k!} \\ &=& \sum_k(-1)^k\frac{k[z^k]\arcsin z}{k!} \\ &=& \sum_k(-1)^k\frac{[z^k]\arcsin z}{(k-1)!} \\ &=& -\sum_m\frac1{(2m)!}\frac{(2m!)}{(m!2^m)^2(2m+1)} \\ &=& -\sum_m\frac1{(m!2^m)^2(2m+1)} \\ &=& -\int_0^1I_0(x)\mathrm dx \\ &=& -\hypergeom12{\frac12}{1,\frac32}{\frac14} \\ &\approx& -1.08652\;, \end{eqnarray} $$

where $I_0$ is the modified Bessel function of the first kind of order $0$ and $F$ is the generalized hypergeometric function.

Using the Taylor series for $(\arcsin z)^2$,

$$(\arcsin z)^2=\sum_{m=0}^\infty\frac{(m!2^m)^2}{(m+1)(2m+1)!}z^{2m+2}\;,$$

the coefficient of $z^{-2}$ is obtained as

$$ \begin{eqnarray} [z^{-1}]z\mathrm e^{-1/\sin z} &=& [z^{-1}]z\sum_k\frac{(-\sin z)^{-k}}{k!} \\ &=&\sum_k(-1)^k\frac{[z^{-2}](\sin z)^{-k}}{k!} \\ &=& \frac12\sum_k(-1)^k\frac{k[z^k](\arcsin z)^2}{k!} \\ &=& \frac12\sum_k(-1)^k\frac{[z^k](\arcsin z)^2}{(k-1)!} \\ &=& \frac12\sum_m\frac1{(2m+1)!}\frac{(m!2^m)^2}{(m+1)(2m+1)!} \\ &=& \frac12\sum_m\frac1{m+1}\left(\frac{m!2^m}{(2m+1)!}\right)^2 \\ &=& \frac\pi2\int_0^1L_0(x)\mathrm dx \\ &=& \frac12\hypergeom23{1,1}{\frac32,\frac32,2}{\frac14} \\ &\approx& 0.528530\;, \end{eqnarray} $$

where $L_0$ is the modified Struve function of order $0$.

Using the Taylor series for $(\arcsin z)^3$,

$$(\arcsin z)^3=6\sum_{m=0}^\infty\frac{(2m+1)!!^2}{(2m+3)!}\sum_{k=0}^m\frac1{(2k+1)^2}z^{2m+3}\;,$$

where a double exclamation mark denotes the double factorial, the coefficient of $z^{-3}$ is obtained as

$$ \begin{eqnarray} [z^{-1}]z^2\mathrm e^{-1/\sin z} &=& [z^{-1}]z^2\sum_k\frac{(-\sin z)^{-k}}{k!} \\ &=&\sum_k(-1)^k\frac{[z^{-3}](\sin z)^{-k}}{k!} \\ &=& \frac13\sum_k(-1)^k\frac{k[z^k](\arcsin z)^3}{k!} \\ &=& \frac13\sum_k(-1)^k\frac{[z^k](\arcsin z)^3}{(k-1)!} \\ &=& -2\sum_m\frac1{(2m+2)!}\frac{(2m+1)!!^2}{(2m+3)!}\sum_{k=0}^m\frac1{(2k+1)^2} \\ &\approx& -0.173756\;. \end{eqnarray} $$

The desired residue is thus given by

$$ \begin{eqnarray} \res{\pi}{z^2\mathrm e^{1/\sin z}} &=& \pi^2\int_0^1\left(L_0(x)-I_0(x)\right)\mathrm dx-2\sum_m\frac1{(2m+2)!}\frac{(2m+1)!!^2}{(2m+3)!}\sum_{k=0}^m\frac1{(2k+1)^2} \\ &=& -\pi^2\hypergeom12{\frac12}{1,\frac32}{\frac14}+\pi\hypergeom23{1,1}{\frac32,\frac32,2}{\frac14} \\ &&-2\sum_m\frac1{(2m+2)!}\frac{(2m+1)!!^2}{(2m+3)!}\sum_{k=0}^m\frac1{(2k+1)^2} \\ &\approx& -7.5764\;, \end{eqnarray} $$

in agreement with your calculation. Here's an evaluation of the result by Wolfram|Alpha, and here's the contour integration.

| cite | improve this answer | |
  • $\begingroup$ Your use of the Lagrange inversion formula here is very clever, and is just what I was looking for. $\endgroup$ – user26872 Mar 15 '12 at 17:41
  • $\begingroup$ @oenamen: Thanks :-). I found that idea in this archived Wikipedia reference desk post, which shows how to calculate the coefficient of $z^{-1}$. $\endgroup$ – joriki Mar 15 '12 at 17:46
  • $\begingroup$ Sometimes it is better to google and then burn paper! $\endgroup$ – user26872 Mar 15 '12 at 17:53
  • 2
    $\begingroup$ @oenamen: I found proofs for the series of $\arcsin^2$ and $\arcsin^3$ here (items 4.3 and 4.4). $\endgroup$ – joriki Apr 3 '12 at 0:41
  • $\begingroup$ Neat and unexpected from the tile (with even an $\,\arcsin^4$ in the previous comment) (+1 of course). $\endgroup$ – Raymond Manzoni May 7 '13 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.