How to Find the Inverse Laplace Transform of arctan(2/s²)

Title: How to Find the Inverse Laplace Transform of arctan(2/s²)

Category: mathematics

The inverse Laplace transform of $\arctan(2/s^2)$ has an elementary closed form, although it is not one of the most basic transform-table pairs. A direct method is to expand the arctangent in powers of $1/s$ and invert each term. This gives an exact series that converges for every finite $t\geq 0$.

Expand the transform

The Taylor series for the arctangent is

$$\arctan x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{2n+1},\qquad |x|<1.$$

Put $x=2/s^2$. For $|s|>\sqrt{2}$, this gives

$$\arctan\left(\frac{2}{s^2}\right) =\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+1}}{2n+1}\frac{1}{s^{4n+2}}.$$

The condition $|s|>\sqrt{2}$ is where this particular series expansion is valid. It is enough to identify the inverse transform, since Laplace transforms are compared in a right half-plane where they converge.

Invert term by term

Use the standard pair

$$\mathcal{L}^{-1}\left\{\frac{1}{s^m}\right\}(t) =\frac{t^{m-1}}{(m-1)!},\qquad m=1,2,3,\ldots.$$

Here $m=4n+2$, so each term contributes a multiple of $t^{4n+1}/(4n+1)!$. Therefore

$$\boxed{ \mathcal{L}^{-1}\left\{\arctan\left(\frac{2}{s^2}\right)\right\}(t) =\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n+1}t^{4n+1}}{(2n+1)(4n+1)!} },\qquad t\geq 0.$$

In fact, this series sums to the elementary expression

$$f(t)=\frac{2\sin(t)\sinh(t)}{t},$$

with the continuous value $f(0)=0$. To verify it, use the identity

$$\mathcal{L}\left\{\frac{\cos(bt)-\cos(at)}{t}\right\}(s) =\frac12\log\left(\frac{s^2+a^2}{s^2+b^2}\right)$$

for sufficiently large real $s$. Taking $a=1+i$ and $b=1-i$, and using $\cos((1-i)t)-\cos((1+i)t)=2i\sin(t)\sinh(t)$, gives

$$\mathcal{L}\left\{\frac{2\sin(t)\sinh(t)}{t}\right\}(s) =\frac{1}{2i}\log\left(\frac{s^2+2i}{s^2-2i}\right) =\arctan\left(\frac{2}{s^2}\right)$$

for sufficiently large real $s$. The series above is the Taylor expansion of this same function.

The first few terms are

$$f(t)=2t-\frac{t^5}{45}+\frac{t^9}{56700}-\frac{128t^{13}}{7\cdot13!}+\cdots.$$

For example, at $t=1$, the first three terms give

$$f(1)\approx 2-\frac{1}{45}+\frac{1}{56700}\approx 1.9777954.$$

The next term changes this approximation by only about $2.94\times10^{-9}$.

Why the series is a valid inverse

The series for the arctangent converges when $|2/s^2|<1$. On the other hand, the Laplace transform of each time-domain term is

$$\mathcal{L}\left\{\frac{(-1)^n2^{2n+1}t^{4n+1}}{(2n+1)(4n+1)!}\right\}(s) =\frac{(-1)^n2^{2n+1}}{(2n+1)s^{4n+2}}.$$

For real $s>\sqrt{2}$, the sum of the absolute values of these transforms converges. Thus the terms can be transformed and summed there, reproducing the arctangent power series. The time-domain series also converges for every finite $t$: the factorial $(4n+1)!$ in its denominator grows faster than the powers in its numerator. These facts verify the transform and yield the stated inverse.

The formula uses the usual principal arctangent on positive real $s$. For complex $s$, branch choices matter; the power-series argument establishes the transform in a right half-plane, which is the relevant setting for the unilateral Laplace transform.

Common mistakes

• Using the wrong arctangent series. The powers are $x^{2n+1}$, so substituting $2/s^2$ produces powers $s^{-(4n+2)}$, not $s^{-(2n+1)}$.
• Forgetting the factorial. The inverse of $1/s^{4n+2}$ is $t^{4n+1}/(4n+1)!$.
• Treating the power series as valid for every $s$. The arctangent expansion requires $|s|>\sqrt{2}$. This restriction applies to that expansion, not to the time-domain series, which converges for all finite $t$.
• Expecting a basic transform-table pair. The answer does have the elementary form $2\sin(t)\sinh(t)/t$, though deriving it from a table may require a logarithmic cosine-difference identity; the series gives a direct alternative.

If you need a numerical value, truncate the series after enough terms for the desired accuracy. For small or moderate $t$, the factorials make the terms decrease quickly; for large $t$, use more terms and check convergence rather than assuming the first few terms suffice.