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I want to determine the value of $s$ for which a particular equation gives the smallest integral with respect to $a$ over the interval $[0,1]$. The three main terms in the equation each involve infinite sums, so the summations are (at least initially) inside the integral of interest.

Fortunately, the summations satisfy the conditions of the dominated convergence theorem so, with the first two terms (not shown here), I'm reasonably confident about being able to bring the integral inside the summation before proceeding to the next stage. However, the last of the three main terms in the equation involves the square of an infinite sum, requiring me to determine $$ \int_0^1\left(\sum_{j=1}^{\infty} \frac{(-1)^j}{j} {\exp ( -j^2 s^2}) \sin (2 j a) \right)^2 \mathrm{d}a $$

Is there anything I can do to make it easier to evaluate this.

Note: the tag is the only one I could think of that was relevant, but please add tag(s) if appropriate

CrimsonDark
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1 Answers1

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The squared expression can be split into a double sum, as follows : $$ \begin{align} \left(\sum_{j=1}^\infty \frac{(-1)^j}{j} e^{-j^2s^2} \sin(2ja)\right)^2 &= \left(\sum_{k=1}^\infty \frac{(-1)^k}{k} e^{-k^2s^2} \sin(2ka)\right) \left(\sum_{l=1}^\infty \frac{(-1)^l}{l} e^{-l^2s^2} \sin(2la)\right) \\ &= \sum_{k,l=1}^\infty \frac{(-1)^{k+l}}{kl} e^{-(k^2+l^2)s^2} \sin(2ka)\sin(2la) \end{align} $$ Then, you will be able to integrate easily $$\sin(2ka)\sin(2la) = \frac{1}{2}(\cos(2(k-l)a)-\cos(2(k+l)a))$$ with respect to $a$. Finally, you may reparametrize the "domain of integration" of the double sum through the change of variables $z_\pm := k \pm l$.

Abezhiko
  • 8,153