a) prove the convergency of $$ \int_0^{\infty} \dfrac {\cosh x \cdot \sin x} {x\cdot e^{x^2}}$$ b) prove the inequality $$ \int_0^{\frac {7\pi} {12}} \dfrac {\cosh x \cdot \sin x} {x\cdot e^{x^2}}<1$$ c) prove the inequality $$ \int_0^{\frac {2\pi} {3}} \dfrac {\cosh x \cdot \sin x} {x\cdot e^{x^2}}>1$$

I have used some Sandor aproximations for b)and c) and even some results of Zitian and Yebing but without success. Any kind of proof is welcome.Thanks!

  • $\begingroup$ I am not sure that $b)$ is correct. $\endgroup$ – Claude Leibovici Aug 25 '15 at 8:28
  • $\begingroup$ i did numerical for bond $\dfrac {\pi} 2$ but is not good for covering the area, the error is to big,wolframalpha gave me the bond of $\dfrac {7\pi} {12}$ did you prove it or just calculated in mathematica $\endgroup$ – Booldy Aug 25 '15 at 8:44
  • $\begingroup$ $b)$ is correct: it is $0.995311$. $\endgroup$ – Patrick Stevens Aug 25 '15 at 14:23
  • 1
    $\begingroup$ $$\int_{0}^{+\infty}e^{-x^2}\cosh(x)\,dx = \frac{1}{2}e^{1/4}\sqrt{\pi},$$ hence convergence follows from $\left|\frac{\sin x}{x}\right|\leq 1$. $\endgroup$ – Jack D'Aurizio Aug 25 '15 at 14:32
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    $\begingroup$ Analytical proofs of inequalities like these that are extremely precise (numerically $0.995 < 1$ and $1 <1.005$) are seldom very illuminating - just careful estimation and alot of work. Anyway we can for example do it with a $30$ order Taylor series of the full integrand, or we can do it with a $4$th order Taylor series of $\frac{\cosh(t)\sin(t)}{t}$ + knowledge of some values of the error-function (which can be found in normal distribution tables if you really want to avoid direct numerics). $\endgroup$ – Winther Sep 12 '15 at 3:12

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