# Homework Help: One hard integral

1. Nov 4, 2011

### matematikuvol

1. The problem statement, all variables and given/known data
Solve

$$\int^{\infty}_0\frac{\sin x}{\sqrt{x}}$$

2. Relevant equations

3. The attempt at a solution
$$\lim_{t\to \infty}\int^{t}_0\frac{\sin x}{\sqrt{x}}$$
$$\sqrt{x}=v$$

so

$$\lim_{t\to \infty}\int^{t}_0\frac{\sin v^2}{v}$$

Integration by parts maybe? What is idea?

Last edited by a moderator: Nov 4, 2011
2. Nov 4, 2011

### ehild

Do not forget dx from the integral. When substituting, you need to transform dx too.

ehild

3. Nov 4, 2011

### matematikuvol

Mistake

I will get

$$2\int^{\infty}_0\sin v^2dv$$

4. Nov 4, 2011

### SammyS

Staff Emeritus
You will get $\displaystyle \int^{t}_0\frac{\sin x}{\sqrt{x}}\,dx=2\int^{\sqrt{t}}_0\sin (v^2)\,dv\,.$

5. Nov 4, 2011

6. Nov 5, 2011

### matematikuvol

But t goes into $$\infty$$ so I write in correct way last step.

7. Nov 5, 2011

8. Nov 5, 2011

### Stimpon

Have you been introduced to the idea of using contours in the complex plane to solve integrals on the real line? Because if you have, that seems like the simplest way to solve this.

9. Nov 5, 2011

### matematikuvol

Is there some other way?

10. Nov 5, 2011

### HallsofIvy

No, not every problem has a trivial solution!

11. Nov 5, 2011

### Stimpon

Nope, have you been introduced to the idea I mentioned? I find it hard to believe that this question would be set if you haven't...

12. Nov 5, 2011

### lurflurf

$$\int^{\infty}_0\frac{\sin x}{\sqrt{x}}=\lim_{a\rightarrow 0^+}\int^{\infty}_0 \frac{\sin x}{\sqrt{x}}e^{-a x}=\lim_{a\rightarrow 0^+}\sqrt{\frac{\pi}{2(a^2+1)(a+\sqrt{a^2+1})}}$$

Complex methods are not the only way to do integrals.
or
The complex part of complex methods to do integrals can be hidden.
Depending on ones perspective.

13. Nov 5, 2011

### I like Serena

I would expect you to have a table of standard integrals or something in which this one is included.

14. Nov 5, 2011

### matematikuvol

And how you get

$$\int^{\infty}_0 \frac{\sin x}{\sqrt{x}}e^{-a x}dx=\sqrt{\frac{\pi}{2(a^2+1)(a+\sqrt{a^2+1})}}$$

15. Nov 6, 2011

### I like Serena

Looks like a Laplace transform.
Seems to me that it's not trivial to calculate it yourself.
Did you use some table to find the transform?

16. Nov 7, 2011

### lurflurf

We want to get rid of the 1/sqrt(t) so substitute
$$\frac{1}{\sqrt{t}}=\frac{2}{\sqrt{\pi}}\int_0^{ \infty } e^{-s^2 t}ds$$
then interchange s and t integrations
$$\int^{\infty}_0 \frac{\sin(b t)}{\sqrt{t}}e^{-a t}dt=\frac{2}{\sqrt{\pi}}\int_0^{ \infty }\left\{\int_0^{\infty}\sin(b t)e^{-a t}e^{-s^2 t}dt \right\}ds=\frac{2}{\sqrt{\pi}}\int_0^{ \infty } \frac{b}{(s^2+a)^2+b^2} ds=b \cdot \sqrt{\frac{\pi}{2} \cdot \frac{1}{(a^2+b^2)(a+\sqrt{a^2+b^2})}}$$
$$\int^{\infty}_0 \frac{\cos(b t)}{\sqrt{t}}e^{-a t}dt= \frac{2}{\sqrt{\pi}}\int_0^{ \infty }\left\{\int_0^{\infty}\cos(b t)e^{-a t}e^{-s^2 t}dt \right\}ds=\frac{2}{\sqrt{\pi}}\int_0^{ \infty } \frac{s^2+a}{(s^2+a)^2+b^2} ds=\sqrt{\frac{\pi}{2} \cdot \frac{a+\sqrt{a^2+b^2}}{a^2+b^2}}$$

That last integral is messy, but it can be done with usual elementary calculus methods and functions. These might be in some Laplace transform tables though often some manipulation is required to get the right form.

Last edited: Nov 7, 2011
17. Nov 7, 2011

### dextercioby

Nice trick, lurflurf. But residue theorem leads to a quicker solution, though. Anyway nice post. :)