# Proving a generalisation of the integral $\int_0^\infty\frac{\sin(x)}{x}dx$

Plugging generalisations of the integral $\int_0^\infty\frac{\sin(x)}{x}dx$ into Wolfram Alpha has lead me to conjecturing the following result:

\begin{align} \int_0^\infty\frac{\sin(x^n)}{x}dx=\frac{\pi}{2n} \tag1 \end{align}

where $n$ is an arbitrary natural number. It holds for values of $n$ from $1$ to $10$, as I've checked using Wolfram Alpha.

However, I haven't had any success computing the integral directly to prove this result. I've seen the case for $n=1$ evaluated using differentiation under the integral sign, so I tried to do that.

We define

\begin{align} I(n,b):=\int_0^\infty\frac{\sin(x^n)}{x}e^{-bx}dx = \Im\int_0^\infty\frac{e^{ix^n-bx}}{x}dx \tag2 \end{align}

Differentiating $(2)$ with respect to $b$, we get

\begin{align} \frac{\partial I(n,b)}{\partial b} = -\Im\int_0^\infty e^{ix^n-bx}dx \tag3 \end{align}

This takes care of the denominator and the negative real part of exponent ensures that the integral converges, but I have no idea how to compute it.

Another approach I tried is to define

\begin{align} I(n):=\int_0^\infty\frac{\sin(x^n)}{x}dx \tag4 \end{align}

and then to differentiate this with respect to $n$, yielding

\begin{align} I'(n) = \int_0^\infty\log(x)x^{n-1}\sin(x^n)dx \tag5 \end{align}

which, using the substitution $z=x^n$, $dz = nx^{n-1}dx$, becomes

\begin{align} I'(n)&=\frac{1}{n}\int_0^\infty\log(\sqrt[n]{z})\sin(z)dz\\ &= \frac{1}{n^2}\int_0^\infty\log(z)\sin(z)dz \tag6 \end{align}

However, $(6)$ is divergent, so it's not useful in computing $I(n)$ either.

Does anybody have any insight into how to make differentiation under the integral sign work for this problem or how to calculate $(3)$, or is it that wrong approach entirely? If it's the wrong approach, how would one go about proving (or disproving, if the whole thing is wrong) my conjectured result?

• Have you tried the substitution $t=x^n$? this gives the answer directly. Dec 21, 2016 at 15:50
• By the change of variable $u=x^n$, $\frac{du}u=n \cdot\frac{dx}x$ one gets $\int_0^\infty\frac{\sin(x^n)}{x}dx=\frac1n\int_0^\infty\frac{\sin(u)}{u}du=\frac{\pi}{2n}$. Dec 21, 2016 at 15:51
• Dammit, how did I miss that?! I thought that if I did that substitution, then $x$ would become $\sqrt[n]{u}$, which didn't make things easier. Guess I should have tried manipulating that subsitution a little more before trying the more complicated methods. At any rate, thanks for pointing that out.
– Tom
Dec 21, 2016 at 15:57

Notice that this question has already been answered in @Oliver Oloa comment: Let $u=x^n$ and the integral trivially relates to the $n=1$ case, which you know how to do.
• And gniourf_gniourf gave it just before Olivier :(. Dec 21, 2016 at 16:46