# Integral inequality

Gold Member
The problem
Show that ## 1-x+ \int^x_1 \frac{\sin t}{t} \ dt < 0## for ## x > 1 ##

The attempt
I rewrite the integral as
##\int^x_1 \frac{\sin t}{t} \ dt < x-1 ##

This is about where I get. Can someone give any suggestions on how to continue from here?

## Answers and Replies

Homework Helper
Gold Member
2021 Award
The problem
Show that ## 1-x+ \int^x_1 \frac{\sin t}{t} \ dt < 0## for ## x > 1 ##

The attempt
I rewrite the integral as
##\int^x_1 \frac{\sin t}{t} \ dt < x-1 ##

This is about where I get. Can someone give any suggestions on how to continue from here?

Let ##f(x) = 1 - x + \int^x_1 \frac{\sin t}{t} \ dt##

• Rectifier
Gold Member
Should I find the derivative to the function and plot it to see if the graph to the function is below the x-axis?

Homework Helper
Gold Member
2021 Award
Should I find the derivative to the function and plot it to see if the graph to the function is below the x-axis?

What's stopping you?

Gold Member
Well, I tried but I stumbled upon this:
##1=\frac{\sin x}{x}## when I derived it and wanted to see where the derivative was 0. I don't think that I can solve that. ;(

Homework Helper
Gold Member
2021 Award
Well, I tried but I stumbled upon this:
##1=\frac{\sin x}{x}## when I derived it and wanted to see where the derivative was 0. I don't think that I can solve that. ;(

What's ##f'(x)##?

Gold Member
It's ##f'(x) = D \left(1-x+ \int^x_1 \frac{\sin t}{t} \ dt \right) = -1+ \frac{\sin x}{x} ##

Homework Helper
Gold Member
2021 Award
And when ##x > 1##?

• Rectifier
Gold Member
Not the same?

Homework Helper
Gold Member
2021 Award
Not the same?

When ##x > 1## then ##f'(x) < 0##. You should be able to see that sort of thing.

• Rectifier
Gold Member
How can I see that? Anything that I should look for? Does it have to do with the fact that the fraction ## \frac{\sin x}{x} ##is always smaller than 1 since ##\lim_{x \rightarrow \infty} \frac{\sin x}{x} = 0##?

EDIT: see edit

Homework Helper
Gold Member
2021 Award
How can I see that? Anything that I should look for? Does it have to do with the fact that the fraction is always smaller than 1?

EDIT: see edit

Yes, ##f(1) = 0##, so if ##f'(x) < 0## for ##x > 1## you are done.

• Rectifier
Gold Member
ok, thanks!