# Solving Hard Integral: \int^{\infty}_0\frac{\sin x}{\sqrt{x}}

• matematikuvol
In summary, the homework statement is to integrate sin x over the range -π to π using the Chain Rule. The Attempt at a Solution suggests that the Integral can be solved using complex methods.

## Homework Statement

Solve

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

## 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?

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

ehild

Mistake

I will get

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

matematikuvol said:
Mistake

I will get

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

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

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

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.

Is there some other way?

No, not every problem has a trivial solution!

matematikuvol said:
Is there some other way?

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...

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

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.

Stimpon said:
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...

matematikuvol said:
Ok. But I would like to see some trick how to calculate this integral in the desert island :)

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

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

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.

And how you get

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

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

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.

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?

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.

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