Integral Stuff 4

Calculus Level 3

$\text{}\int_{-\infty }^{\infty } \sin \left(e^x\right) \, dx=\pi / a$ . Find a.

2 solutions

D S
Nov 16, 2017

Take x=lnu and the integral becomes 0 to infinity sinx/x so 2

Munem Shahriar
Nov 16, 2017

First, computing the indefinite integral:

$\int \sin \left(e^x\right) dx$

$= \int \dfrac{\sin (u)}{u} du+ C$ $~ ~ ~ ~ ~ ~ ~ ~ ~~ ~$ $;[ u = e^x]$

$= Si (u) + C$

$= Si (e^x) + C$

Now, computing the boundaries

$\lim_{x \to -\infty} (Si (e^x)) = 0$

$\lim_{x \to \infty} (Si (e^x) ) = \dfrac{\pi}{2}$

Hence,

$\int_{-\infty }^{\infty } \sin \left(e^x\right) dx$

$= \dfrac{\pi}{2} -0$

$= \dfrac{\pi}{2}$

Hece $a = \boxed{2}$

Most readers are interested in how to find the primitive of that sine. Not the answer, Amy possibility you can explain that part?

Peter van der Linden - 3 years, 6 months ago

This solution is incomplete. You didn't explain how you move from the first line to the second line.

Pi Han Goh - 3 years, 6 months ago

It should be clearer now.

Munem Shahriar - 3 years, 6 months ago

You didn't explain the most important step!

How did you show that $\displaystyle \int_{-\infty}^\infty \sin(e^x) \, dx = \frac\pi2 - 0$ in the first place?!

Pi Han Goh - 3 years, 6 months ago

@Pi Han Goh – Is that okay?

Munem Shahriar - 3 years, 6 months ago

@Munem Shahriar – That's better. But you still missed out a crucial step:

How do you prove that $\displaystyle \int_0^{\infty} \dfrac{\sin x}x \, dx = \dfrac\pi 2$ ?

Pi Han Goh - 3 years, 6 months ago

@Pi Han Goh – i don't think @Munem Sahariar has to prove that as it is a basic step of integration.

Mohammad Khaza - 3 years, 6 months ago