$$I_1=\int_{0}^{\infty}\frac{\sin(x)}{\sinh(x)}dx,$$ $$I_2=\int_{0}^{\infty}\frac{\sin(x)}{\cosh(x)}dx,$$ $$I_3=\int_{0}^{\infty}\frac{\cos(x)}{\cosh(x)}dx.$$

Which of the following is true?

$\color{blue}{\text{(A)}\space\space\space I_3<I_2<I_1✓}$

$\text{(B)}\space\space\space I_3<I_1<I_2$

$\text{(C)}\space\space\space I_2<I_3<I_1$

$\text{(D)}\space\space\space I_2<I_1<I_3$

$\text{(E)}\space\space\space I_1<I_3<I_2$

Is there a clever way to answer this question in less than $4$ minutes and without using a calculator?

I have solved it correctly, but in a very long time that is not suitable for an MCQ exam.

I believe there is something obvious for some of you, or a way without evaluating.

Your help would be appreciated. THANKS!

  • 1
    $\begingroup$ You can get that $I_2<I_1$ almost immediately by just noticing that $\cosh(x)>\sinh(x)$ by looking at their definitions. $\endgroup$
    – Lorago
    Feb 6, 2022 at 16:42
  • $\begingroup$ @Lorago Thank you for this is a good information. We therefore exclude options B and E. Now we need another idea to exclude options C and D. $\endgroup$ Feb 7, 2022 at 10:38

1 Answer 1


Sorry, too long for a comment

I'm not sure whether this task can be rigorously done in 4 minutes. But heuristically it can be done. First, we note that the integrands decline exponentially, and at $x\to\pi \,\,\,|\frac{1}{\cosh \pi}|\sim|\frac{1}{\sinh \pi}|\sim 0.1$, so the integrands decline 10 times. Therefore, for comparison we can consider the integration from $0$ to $\pi$. On the interval $[0;\pi] \,\, \frac{\sin x}{\sinh x}>\frac{\sin x}{\cosh x}$ (because $\cosh x>\sinh x$, as @Lorago mentioned), so $I_1>I_2$. Next, let's integrate $I_3$ by part: $\int_{0}^{\infty}\frac{\cos x }{\cosh x }dx=\frac{\sin x}{\cosh x}\Big|_0^\infty+\int_{0}^{\infty}\frac{\sin x }{\cosh x }\frac{\sinh x}{\cosh x}dx$. On the interval $[0;\pi] \,\, \frac{\sin x }{\cosh x }\frac{\sinh x}{\cosh x}<\frac{\sin x }{\cosh x }$, so $I_3<I_2$


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