# Integral inequality

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1. Jan 5, 2017

### Rectifier

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?

2. Jan 5, 2017

### PeroK

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

3. Jan 5, 2017

### Rectifier

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

4. Jan 5, 2017

### PeroK

What's stopping you?

5. Jan 5, 2017

### Rectifier

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. ;(

6. Jan 5, 2017

### PeroK

What's $f'(x)$?

7. Jan 5, 2017

### Rectifier

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

8. Jan 5, 2017

### PeroK

And when $x > 1$?

9. Jan 5, 2017

### Rectifier

Not the same?

10. Jan 5, 2017

### PeroK

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

11. Jan 5, 2017

### Rectifier

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

12. Jan 5, 2017

### PeroK

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

13. Jan 5, 2017

ok, thanks!