Proof to $\int_{0}^{\infty}\sin(t)t^{z-1}\,\mathrm{d}t= \sin\left ( \frac{\pi z}{2} \right )\Gamma(z)$

Question

I tried to check the source of the proof to the equation $$\int_{0}^{\infty}\sin(t)t^{z-1}\,\mathrm{d}t= \sin\left ( \frac{\pi z}{2} \right )\Gamma(z),\qquad -1<\Re z < 1$$ but it only has a sketch of proof, since it was meant to be an exercise. In this exercise seems a bit unclear to me what to do. I am looking for a proof to this equation, if there is one. Another source says the domain should be $0<\Re z < 1$ (in which it has no proof). I do not know what this equation is called so searching on Google is kind of difficult.

Answer 1

For $\text{Re}(z)\in(-1,1)$, we may use the Laplace transform to get:

$$ \int_{0}^{+\infty}\sin(t)\,t^{z-1}\,dt = \frac{1}{\Gamma(1-z)}\int_{0}^{+\infty}\frac{ds}{s^z(1+s^2)}\tag{1}$$ since $\mathcal{L}(\sin t)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(t^{z-1}\right)=\frac{1}{s^z \Gamma(1-z)}.$

With the substitution $\frac{1}{1+s^2}=u$, the RHS of $(1)$ becomes: $$ \frac{1}{2\,\Gamma(1-z)}\int_{0}^{1}u^{\frac{z-1}{2}}(1-u)^{\frac{-z-1}{2}}\,du=\frac{\Gamma\left(\frac{z+1}{2}\right)\,\Gamma\left(\frac{1-z}{2}\right)}{2\,\Gamma(1-z)} \tag{2}$$ by Euler's beta function, then the claim follows from the $\Gamma$ reflection formula: $$ \Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}.\tag{3}$$

Answer 2

Let $I(s)$ be the integral given by

$$I(s)=\int_0^\infty t^{s-1}\sin(t)\,dt \tag 1$$

for $\text{Re}(s)<1$.

We can use Euler's Formula to write $(1)$ as

$$I(s)=\frac1{2i}\left(\int_0^\infty t^{s-1}e^{it}\,dt-\int_0^\infty t^{s-1}e^{-it}\,dt\right) \tag 2$$

Now, moving to the complex plane, we analyze the closed contour integral(s)

$$J(s)=\oint_{C^\pm} z^{s-1}e^{\pm iz}\,dz \tag 3$$

where $C^\pm$ is comprised of (i) the segment on the real line from $\epsilon$ to $R$, (ii) the quarter circle with radius $R$, centered at the origin, from $R$ to $\pm iR$, (iii) the line segment along the imaginary axis from $\pm iR$ to $\pm i\epsilon$, and (iv) the quarter circle with radius $\epsilon$, centered at the origin, from $ \pm i\epsilon$ to $\epsilon$.

With the branch cut chosen as the line along the non-positive real axis, from $0$ to $-\infty$, the integrands in $3$ are analytic in and on $C^\pm $. Then, Cauchy's Integral Theorem guarantees that $J(s)=0$.

In the limit as $R\to \infty$ and $\epsilon \to 0$, the contributions to $J(s)$ from the integrations along the quarter circles vanish. Hence, we find that

$$\begin{align} \int_0^\infty t^{s-1}e^{\pm i t}\,dt&=(\pm i)^{s}\int_0^\infty t^{s-1}e^{-t}\,dy\\\\ &=e^{\pm i\pi s/2}\Gamma(s)\tag 4 \end{align}$$

Using $(4)$ in $(2)$, we find

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^{s-1}\sin(t)\,dt =\sin(\pi s/2)\Gamma(s)}$$

as was to be shown!