# Prove the existence of $t_{0}$ such that $\int_{0}^{t}\frac{\sin{x}}{\sqrt{x}}dx+\sin{t}>0$ for all $t\geq t_{0}$

I am trying to prove the existence of $t_{0}$ such that $$\int_{0}^{t}\frac{\sin{x}}{\sqrt{x}}dx+\sin{t}>0$$ for all $t\geq t_{0}$.

Numerical experiments shows this (see the figure). The first thing that may come to mind is to show the existence of $t_{0}$ such that $$\int_{0}^{t}\frac{\sin{x}}{\sqrt{x}}dx>1$$ when $t\geq t_{0}$

• sin(t) = the integral from 0 to t of -cos(t) from 0 to t , then combine the 2 integrals . – StuartMN Sep 18 '18 at 1:26
• Nonsense ,sin(0)=0 – StuartMN Sep 18 '18 at 1:40
• Yeah, sorry. I was just about to correct my comment. But how does that simplify ? – Medo Sep 18 '18 at 1:41

## 2 Answers

The full (improper Riemann) integral on the half line is well known, see Proof of $\int_0^\infty \frac{\sin x}{\sqrt{x}}dx=\sqrt{\frac{\pi}{2}}$ . (A very related term to google : Fresnel integral.) The definition of having a finite limit as $t\to\infty$ immediately gives the existence of some $t_0>0$ such that if $t>t_0$,

$$\left| \int_0^t \frac{\sin x}{\sqrt x}\ dx - \sqrt{\frac{\pi}{2}} \right| < 0.1.$$

As $\sqrt{\pi/2} > 1.2$, $\int_0^t \frac{\sin x}{\sqrt x}\ dx > 1.1 > 1$ for $t>t_0$, as required.

• Just brilliant... – Medo Sep 18 '18 at 2:51
• @Medo you're welcome :) – Calvin Khor Sep 18 '18 at 2:51

Let $t>\pi$. Then

$$\left|\int_{5\pi}^{t}\frac{\sin{x}}{\sqrt{x}}dx\right|\leq \frac{1}{\sqrt{5\pi}} \left|\int_{5\pi}^{t}\sin{x}dx\right| \leq \frac{2}{\sqrt{5\pi}}=0.5046265$$

Using matlab, $$\int_{0}^{5\pi}\frac{\sin{x}}{\sqrt{x}}dx=1.5048853$$

Therefore, if $t>\pi$ then $\int_{5\pi}^{t}\frac{\sin{x}}{\sqrt{x}}dx>1$.

Now, can it be done analytically ?