Convergence of an alternating improper integral with a parameter I have the following question:
Let $\alpha \in \Bbb R$ be a parameter.
Study the convergence of
$$ \int_{-\infty}^\infty \frac{\sin(e^x)}{e^{\alpha x}+ e^{(\alpha +1) x}}dx$$
I split the integral in two parts:
$$ \int_{-\infty}^0 \frac{\sin(e^x)}{e^{\alpha x}+ e^{(\alpha +1) x}}dx  $$
and
$$  \int_{0}^\infty \frac{\sin(e^x)}{e^{\alpha x}+ e^{(\alpha +1) x}}dx $$
For the first one, one can simply use the asymptotic equivalence relation $\sin(e^x) \sim e^x$ when $x\to-\infty$ , thus yielding to the necessary and sufficient condition for convergence of $\alpha \lt 1$.
For the second integral, I approached it using the Abel-Dirichlet test mentioned here, and I concluded that, if $\alpha \gt -2$, then the integral converges.
However, $\alpha \gt -2$ is only a sufficient condition for the convergence, so I can't really tell if there exists some $\alpha \le -2$ such that the hypothesis for the Abel-Dirichlet criterion are not fulfilled but the integral still converges, and thus the solution is incomplete.
How could this problem be approached? I thought about different things but all of them seem not to work.
And, in general, I'd be very grateful if someone could explain to me how to approach the study of improper alternating integrals with a parameter, since the elementary results about non-negative (or non-positive) functions do not work here and even the Abel-Dirichlet test is not suitable for a problem with a parameter.
Thank you.
 A: Let $t=e^x$ then we have
$$\int_0^\infty \frac{\sin t}{1+t}\frac{dt}{t^{\alpha+1}}$$
It should be clear the integral diverges near $0$ when $\alpha \geq 1$. For other values that would normally cause a divergence at $\infty$, we can use the alternating series test
$$ = \sum_{k=0}^\infty \int_{k\pi}^{(k+1)\pi} \frac{\sin t}{1+t}\frac{dt}{t^{\alpha+1}}$$
to conclude the series converges as long as the absolute value of the integral of the integrand between the two nodes approaches $0$, which doesn't happen when $\alpha \leq -2$. Thus the integral converges for $ -2 < \alpha < 1$
A: @Ninad Munshi gave the right explanations.
For $\alpha=0$, the result is
$$ \text{Si}(1) \cos (1)- \text{Ci}(1) \sin (1)+\frac \pi 2(1-\cos(1))$$
What is interesting is the case where $\alpha$ is the reciprocal of an integer. Let $\alpha=\frac 1n$. We obtain simple closed form expression for $n=\pm 2$.
For $\alpha=\frac 12$
$$\sqrt{2 \pi }-\pi  \sin (1) \left(C\left(\sqrt{\frac{2}{\pi }}\right)+S\left(\sqrt{\frac{2}{\pi
   }}\right)-1\right)+$$ $$\pi  \cos (1) \left(S\left(\sqrt{\frac{2}{\pi
   }}\right)-C\left(\sqrt{\frac{2}{\pi }}\right)\right)$$
For $\alpha=-\frac 12$
$$\pi  \left(\sin (1) \left(C\left(\sqrt{\frac{2}{\pi }}\right)+S\left(\sqrt{\frac{2}{\pi
   }}\right)-1\right)+\cos (1) \left(C\left(\sqrt{\frac{2}{\pi
   }}\right)-S\left(\sqrt{\frac{2}{\pi }}\right)\right)\right)$$  For other values, arrive some nasty hypergeomtric functions.
