Evaluating an integral using the gamma function

Question

My question regards an integral

$$\int_0^\infty \frac{\sin(x^p)}{x^p}\mathrm{d}x$$ The answer should be $$\frac{1}{p-1}\cos(\frac{\pi}{2p})\Gamma(\frac{1}{p})$$ and I roughly know that I should apply some kind of substitution and the Cauchy theorem, but can't seem to find a way to gain this solution, I am stuck at choosing the right substitution.

Answer 1

First, the obvious substitution is:

$$t=x^p$$

Not thinking about the conditions for convergence for now, we have the integral:

$$I=\int_0^\infty \frac{\sin(x^p)}{x^p}\mathrm{d}x=\frac{1}{p}\int_0^\infty \frac{\sin(t)}{t^{2-1/p} }\mathrm{d}t$$

Now let's put $2-1/p=q$.

The trick is to turn this into a double integral. Notice that:

$$\int_0^{\infty} u^{q-1} e^{-tu}\mathrm{d}u=\frac{\Gamma(q)}{t^q}$$

So we have:

$$I=\frac{1}{p~ \Gamma(q)}\int_0^\infty \int_0^\infty u^{q-1} \sin(t) e^{-tu}\mathrm{d}u ~\mathrm{d}t$$

Now we reverse the order of integration and take the following integral:

$$I_1=\int_0^\infty \sin(t) e^{-tu}\mathrm{d}t$$

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This integral is easy to solve integrating by parts two times:

$$I_1=-\sin t \frac{1}{u} e^{-tu}\bigg|_0^\infty+\frac{1}{u}\int_0^\infty \cos(t) e^{-tu}\mathrm{d}t=\frac{1}{u}\int_0^\infty \cos(t) e^{-tu}\mathrm{d}t=$$

$$=-\frac{1}{u^2} \cos (t)~ e^{-tu} \bigg|_0^\infty-\frac{1}{u^2} \int_0^\infty \sin(t) e^{-tu}\mathrm{d}t=\frac{1}{u^2}(1-I_1)$$

$$\left( 1+\frac{1}{u^2}\right)I_1=\frac{1}{u^2}$$

$$I_1=\frac{1}{1+u^2}$$

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From this our doulbe integral will turn into:

$$I=\frac{1}{p~\Gamma(q)}\int_0^\infty \frac{u^{q-1}}{1+u^2} \mathrm{d}u $$

Let's put $u^2=v$ and get:

$$I=\frac{1}{2 p~\Gamma(q)}\int_0^\infty \frac{v^{q/2-1}}{1+v} \mathrm{d}v $$

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From one of the definitions of the Beta function:

$$\int_0^{\infty} \frac{x^{a-1}}{1+x}dx=B(a,1-a),~~~~1>a>0$$

By the properties of the Beta function:

$$B(a,b)=\frac{\Gamma(a) \Gamma(b)}{\Gamma(a+b)}$$

$$B(a,1-a)=\frac{\Gamma(a) \Gamma(1-a)}{\Gamma(1)}=\Gamma(a) \Gamma(1-a)$$

By the reflection formula for the Gamma function:

$$\Gamma(a) \Gamma(1-a)=\frac{\pi}{\sin (\pi a)}$$

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Here:

$$a=\frac{q}{2}=\frac{2-1/p}{2}=1-\frac{1}{2p}$$

For $1>a>0$ we have the condition $p>\frac{1}{2}$.

$$\sin (\pi a)=\sin \left(\pi -\frac{\pi}{2p} \right)=\sin \left(\frac{\pi}{2p} \right)$$

Also by the main property of Gamma function and by the reflection formula:

$$\Gamma(q)=\Gamma \left(2-\frac{1}{p} \right)=\left( 1-\frac{1}{p} \right) \Gamma \left(1-\frac{1}{p} \right)=\frac{p-1}{p}\frac{\pi}{\sin (\pi/p) \Gamma(1/p)}$$

Finally, we obtain:

$$I=\frac{1}{2 p} \frac{p} {p-1} \frac{\sin (\pi/p) \Gamma(1/p)}{\pi} \frac{\pi}{\sin \left( \pi/(2p) \right)}$$

And the answer is indeed:

$$I=\frac{1}{p-1}\cos \left( \frac{\pi}{2p} \right) \Gamma \left( \frac{1}{p} \right)$$

Answer 2

Once the integral is in the form $$ I_p = \frac{1}{p}\int_{0}^{+\infty}\frac{\sin x}{x^{2-1/p}}\,dx \tag{1}$$ we may use the Laplace transform, giving $$ \mathcal{L}(\sin x)=\frac{1}{1+s^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{x^{2-1/p}}\right)=\frac{1}{\Gamma\left(2-\frac{1}{p}\right)}\,s^{1-1/p}\tag{2}$$ and $$ I_p = \frac{1}{p\,\Gamma\left(2-\frac{1}{p}\right)}\int_{0}^{+\infty}\frac{s^{1-1/p}}{1+s^2}\,ds \tag{3}$$ is converted into an Euler beta integral by the substitution $\frac{1}{1+s^2}=u$.

The final outcome is a consequence of the $\Gamma$ reflection formula $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$

Answer 3

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\begin{align} \color{#f00}{\int_{0}^{\infty}{\sin\pars{x^{p}} \over x^{p}}\,\dd x} &\ \stackrel{x^{p}\ \to\ x}{=}\ {1 \over p}\int_{0}^{\infty}x^{1/p - 1}\,{\sin\pars{x} \over x}\,\dd x = {1 \over p}\int_{0}^{\infty}x^{1/p - 1}\ \overbrace{\int_{0}^{1}\cos\pars{kx}\,\dd k}^{\ds{{\sin\pars{x} \over x}}}\ \,\dd x \\[3mm] & = {1 \over p}\,\Re\int_{0}^{1} \int_{0}^{\infty}x^{1/p - 1}\expo{\ic kx}\,\dd x\,\dd k = {1 \over p}\,\Re\int_{0}^{1}k^{-1/p} \int_{0}^{\infty}x^{1/p - 1}\expo{\ic x}\,\dd x\,\dd k = \\[3mm] & = {1 \over p -1}\,\Re\int_{0}^{\infty}x^{1/p - 1}\expo{\ic x}\,\dd x\tag{1} \end{align}

The '$x$-integral' is 'closed' with a quarter arc above the complex plane and the path $\braces{z = y\ic\ |\ y > 0}$. Integration along the arc vanishes out. The $x$-integration is performed by setting the branch-cut

$$ z^{1/p - 1} = \verts{z}^{1/p - 1}\exp\pars{\ic\bracks{{1 \over p} - 1} \mathrm{arg}\pars{z}}\,,\quad \verts{\mathrm{arg}\pars{z}} < \pi\,,\quad z \not= 0 $$.

Namely ( with the result $\pars{1}$ ) \begin{align} \color{#f00}{\int_{0}^{\infty}{\sin\pars{x^{p}} \over x^{p}}\,\dd x} & = {1 \over p - 1}\,\Re\bracks{-\int_{\infty}^{0}y^{1/p - 1} \exp\pars{\ic\,{\pi \over 2}\bracks{{1 \over p} - 1}}\expo{-y}\,\ic\dd y} \\[3mm] & = {1 \over p - 1}\,\Re\pars{\exp\pars{{\pi \over 2p}\,\ic}} \int_{0}^{\infty}y^{1/p - 1}\expo{-y}\,\dd y \\[3mm] & = \color{#f00}{{1 \over p - 1}\,\cos\pars{{\pi \over 2p}}\Gamma\pars{{1 \over p}}} \end{align}