# Arranging $\int_{0}^{\infty}\frac{\sin(x)}{\sinh(x)}dx, \int_{0}^{\infty}\frac{\sin(x)}{\cosh(x)}dx, \int_{0}^{\infty}\frac{\cos(x)}{\cosh(x)}dx$

Let

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

Which of the following is true?

$$\color{blue}{\text{(A)}\space\space\space I_3

$$\text{(B)}\space\space\space I_3

$$\text{(C)}\space\space\space I_2

$$\text{(D)}\space\space\space I_2

$$\text{(E)}\space\space\space I_1

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!

• You can get that $I_2<I_1$ almost immediately by just noticing that $\cosh(x)>\sinh(x)$ by looking at their definitions. Feb 6, 2022 at 16:42
• @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. Feb 7, 2022 at 10:38

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