1^()circ Show I'(a)=dfrac a2 sum from k=1 to infinity
$1^\circ$ Show $I'(a)=\dfrac a2\displaystyle\sum_{k=1}^{\infty}\dfrac1{k^2+a^2}$. $2^\circ$ Using the partial-fraction expansion $\coth x=\dfrac1x+\displaystyle\sum_{n=1}^{\infty}\dfrac{2x}{x^2+n^2\pi^2}$, find $I(a)$ in elementary terms.
Explanation of Method
Differentiating under the integral sign turns the $\sin^2$ into $\sin(2ax)$; expanding $1/(e^{2x}-1)$ as a geometric series and using the standard Laplace transform of $\sin$ reduces $I'(a)$ to a simple sum over $k$, which the given $\coth$ identity evaluates in closed form.
- Differentiate under the integral sign and expand $1/(e^{2x}-1)$.
- Evaluate each term via the Laplace transform of $\sin$.
- Apply the $\coth$ identity, then integrate back and fix the constant.
Part 1° · Differentiate under the integral
Since $\dfrac{d}{da}[\sin^2(ax)]=x\sin(2ax)$: $$ I'(a) = \int_0^{\infty}\frac{\sin(2ax)}{e^{2x}-1}\,dx. $$ Expanding $\dfrac1{e^{2x}-1}=\sum_{k=1}^{\infty}e^{-2kx}$ (geometric series, valid for $x>0$) and using the Laplace transform $\int_0^{\infty}e^{-pt}\sin(qt)\,dt=\dfrac q{p^2+q^2}$ with $p=2k$, $q=2a$: $$ I'(a) = \sum_{k=1}^{\infty}\frac{2a}{4k^2+4a^2} = \frac a2\sum_{k=1}^{\infty}\frac1{k^2+a^2}. $$
Part 2° · Apply the coth identity
Setting $x=\pi a$ in $\coth x=\dfrac1x+\sum_{n=1}^{\infty}\dfrac{2x}{x^2+n^2\pi^2}$ and simplifying each term ($\dfrac{2\pi a}{\pi^2a^2+n^2\pi^2}=\dfrac{2a}{\pi(a^2+n^2)}$): $$ \pi\coth(\pi a) = \frac1a + 2a\sum_{n=1}^{\infty}\frac1{n^2+a^2} \implies \sum_{k=1}^{\infty}\frac1{k^2+a^2} = \frac{\pi\coth(\pi a)-1/a}{2a}. $$ Substituting into $I'(a)$: $$ I'(a) = \frac a2\cdot\frac{\pi\coth(\pi a)-1/a}{2a} = \frac14\left[\pi\coth(\pi a)-\frac1a\right]. $$
Step 3 · Integrate and fix the constant
$$ I(a) = \frac14\left[\ln\sinh(\pi a) - \ln a\right] + C = \frac14\ln\frac{\sinh(\pi a)}a + C. $$ Since $\sinh(\pi a)/a\to\pi$ as $a\to0$, and $I(0)=0$ (the original integrand vanishes identically at $a=0$): $$ 0 = \frac14\ln\pi + C \implies C = -\frac14\ln\pi. $$
Summary Table
| Property | Value |
|---|---|
| $I'(a)$ | $\tfrac14[\pi\coth(\pi a)-1/a]$ |
| $I(a)$ | $\tfrac14\ln\dfrac{\sinh(\pi a)}{\pi a}$ |
| $I(0)$ | $0$ |
Additional Notes
- As $a\to0$, $\sinh(\pi a)/(\pi a)\to1$ smoothly, so $I(a)\to0$ continuously — consistent with the original integrand $\sin^2(ax)$ vanishing identically at $a=0$.
- The Mittag-Leffler partial-fraction expansion of $\coth x$ used here is the same type of identity that generates the classical formula $\pi\cot(\pi z)=1/z+\sum2z/(z^2-n^2)$ for the ordinary cotangent, just with hyperbolic functions instead of trigonometric ones.
- As $a\to\infty$, $\coth(\pi a)\to1$ and $\sinh(\pi a)\sim e^{\pi a}/2$, so $I(a)\sim\frac{\pi a}4-\frac14\ln(2\pi a)$, growing essentially linearly — reflecting how the oscillation frequency $a$ increasingly dominates the exponentially-weighted denominator.