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I have a problem

$$\int_{0}^{+\infty} \frac{\sin^3 x}{x^2} \,dx$$

How do I solve this?

Attempt:

$$I(\alpha)=\int_{0}^{+\infty} \frac{\sin^3 (\alpha x)}{x^2} \,dx$$

$$\implies I'(\alpha)=\int_{0}^{+\infty} \frac{3\cos (\alpha x)\sin^2 (\alpha x)}{x} \,dx$$

$$\Longleftrightarrow I'(\alpha)=\frac{3}{2}\int_{0}^{+\infty} \frac{\sin (\alpha x)\sin (2\alpha x)}{x} \,dx$$

$$\Longleftrightarrow I'(\alpha)=\frac{3}{2}\int_{0}^{+\infty} \frac{\cos (\alpha x)-\cos (3\alpha x)}{x} \,dx$$

I'm stuck here. If I continue to differential it one more time I get

$$I''(\alpha)=\frac{3}{2}\int_{0}^{+\infty} -\sin (\alpha x)+3\sin (3\alpha x) \,dx$$

In which I also don't know what to proceed next. Thanks everyone!

user635988
  • 858

1 Answers1

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There is an error with the last expression for $I'(\alpha)$. It should read

$$I'(\alpha)=\frac34 \int_0^\infty \frac{\cos(\alpha x)-\cos(3\alpha x)}{x}\,dx$$

Now, using an extension of Frullani's integral (See HERE, HERE, and HERE), we have

$$\int_0^\infty \frac{\cos(\alpha x)-\cos(3\alpha x)}{x}\,dx=\log(3)$$

Then, $I'(\alpha)=\frac34\log(3)$, which implies that $I(\alpha)=\frac34\log(3)\alpha+C$. And since $I(0)=0$, then $C=0$ and we are done!

$$\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx=\frac34\log(3)$$

NOTE:

Rather than appeal to Feynman's trick, we could use integration by parts with $u=\sin^3(x)$ and $v=-\frac1x$ to find that

$$\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx=\frac34 \int_0^\infty \frac{\cos(x)-\cos(3x)}{x}\,dx$$

Applying the extended version of Frulani's integral, we arrive at the result

$$\int_0^\infty \frac{\sin^3(x)}{x^2}\,dx=\frac34\log(3)$$

as expected!

Mark Viola
  • 179,405