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I highly suspect that $$\int_{0}^{\infty} \cos(\cosh x) \cosh (\alpha x) \, \mathrm dx$$ converges for $0\le \alpha <1$

(If true, it obviously also converges for $-1 < a <0$.)

I can show that the integral converges for $\alpha=0$: $$\int_{0}^{\infty} \cos(\cosh x) \, \mathrm dx= \int_{1}^{\infty} \frac{\cos (u)}{\sqrt{u^{2}-1}} \, \mathrm du$$ which converges by Dirichlet's test

I can also show that the integral doesn't converge for $\alpha=1$:

$$\int_{0}^{\infty} \cos(\cosh x) \cosh(x) \, \mathrm dx = \int_{1}^{\infty} \frac{u \cos (u)}{\sqrt{u^{2}-1}} \, \mathrm du$$ which doesn't converge since $\frac{u \cos (u)}{\sqrt{u^{2}-1}} \sim \cos (u)$ for large values of $u$

For other values of $\alpha$ between $0$ and $1$, I'm not sure what to do. I don't know how to express $\cosh (\alpha x)$ in terms of $\cosh (x)$.

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For $\alpha \in [0, 1)$, let $t = e^x$. Then

$$ \int_{0}^{R} \cos(\cosh x)\cosh(\alpha x) \, dx = \int_{1}^{e^R} \cos\left( \frac{t+t^{-1}}{2} \right)\frac{t^{\alpha}+t^{-\alpha}}{2t} \, dt. $$

Noticing that

$$ \cos\left( \frac{t+t^{-1}}{2} \right) = \cos\left( \frac{t}{2} \right) + \mathcal{O}\left( \frac{1}{2t} \right) \quad \text{as } t\to\infty, $$

we find that the integral converges as $R\to\infty$ by Dirichlet's test.

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  • $\begingroup$ I was going to ask you about that expansion, but then I realized it follows almost immediately from the trig identity $\cos(x+y) = \cos(x) \cos(y) - \sin (x) \sin (y)$. $\endgroup$ – user232456 Oct 14 '17 at 0:32
  • $\begingroup$ @user232456, That's true. Alternatively you may count on the mean value theorem :) $\endgroup$ – Sangchul Lee Oct 14 '17 at 1:06
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$\cosh^{-1}u = \ln(u+\sqrt{u^2-1})$, so $\cos(\alpha\cosh^{-1}(u))=\frac12\big((u+\sqrt{u^2-1})^\alpha+(u-\sqrt{u^2-1})^\alpha\big)$.

$$\int_{0}^{\infty} \cos(\cosh x) \cosh(\alpha x) \, \mathrm dx = \int_{1}^{\infty} \frac{\cosh(\alpha\cosh^{-1}u) \cos (u)}{\sqrt{u^{2}-1}} \, \mathrm du \sim \int_{1}^{\infty} \frac1{u^{1-\alpha}}\cos (u)\, \mathrm du.$$ So the integral converges for $\alpha\in[0,1)$ and diverges for $\alpha\in[1,\infty)$.

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