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Prove that:

$$\int_{0}^{\infty}t^a e^{-t}\sin t \,dt=\frac{a!}{2^{(a+1)/2}}\sin \frac{\left ( a+1 \right )\pi}{4}, \;\;\; a \in \mathbb{Z}_{+}$$

Using integration by parts successively, we have$^{*}$ that\begin{align*}
\left.\begin{array}{l}
J_1=\displaystyle\int_{0}^{+\infty}{x^a\,{\rm{e}}}^{(-1+i)\,x}\,dx=a!\,(1-i)^{-1-a}\\
J_2=\displaystyle\int_{0}^{+\infty}{x^a\,{\rm{e}}}^{(-1-i)\,x}\,dx=a!\,(1+i)^{-1-a}
\end{array}\,,\quad a\in\mathbb{Z}^{+}\right\}
\end{align*} Then
\begin{align*}
\displaystyle\int_{0}^{+\infty}{x^a\,{\rm{e}}}^{-x}\sin{x}\,dx&=\frac{1}{2i}\,(J_1-J_2)\\
&=\frac{a!}{2i}\,\big((1-i)^{-1-a}-(1+i)^{-1-a}\big)\\
&=\frac{a!}{2^{\frac{a+1}{2}}\,2i}\,\bigg(\Big(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\,i\Big)^{-1-a}-\Big(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\,i\Big)^{-1-a}\bigg)\\
&=\frac{a!}{2^{\frac{a+1}{2}}\,2i}\,\Big(\cancel{\cos\tfrac{(a+1)\,\pi}{4}}+i\,\sin\tfrac{(a+1)\,\pi}{4}-\cancel{\cos\tfrac{(-a-1)\,\pi}{4}}-i\,\sin\tfrac{(-a-1)\,\pi}{4}\Big)\\
&=\frac{a!}{2^{\frac{a+1}{2}}\,2i}\,2i\,\sin\tfrac{(a+1)\,\pi}{4}\\
&=\frac{a!}{2^{\frac{a+1}{2}}}\,\sin\tfrac{(a+1)\,\pi}{4}\,.\end{align*}



$\begin{aligned}
(*)\quad &\mathop{\lim}\limits_{x\to+\infty}x^a\,{\rm{e}}^{(-1+i)x}=0\\
&\mathop{\lim}\limits_{x\to+\infty}x^a\,{\rm{e}}^{(-1-i)x}=0 \end{aligned}$

Grigorios Kostakos wrote:Using integration by parts successively, we have$^{*}$ that\begin{align*}
\left.\begin{array}{l}
J_1=\displaystyle\int_{0}^{+\infty}{x^a\,{\rm{e}}}^{(-1+i)\,x}\,dx=a!\,(1-i)^{-1-a}\\
J_2=\displaystyle\int_{0}^{+\infty}{x^a\,{\rm{e}}}^{(-1-i)\,x}\,dx=a!\,(1+i)^{-1-a}
\end{array}\,,\quad a\in\mathbb{Z}^{+}\right\}
\end{align*}

This step can be made easily invoking a Laplace. For example if we recall that

$$\int_{0}^{\infty} x^n e^{-st} \, {\rm d}t = \frac{\Gamma(n+1)}{s^{n+1}} \tag{1}$$

Now what we want follows directly if we transform $(1)$ as follows: $n \mapsto a$, $s \mapsto 1-i$ and $s \mapsto -1-i$ respectively. For example for the first integral we have that:

\begin{align*}
\int_{0}^{\infty} x^a e^{(-1+i)x} \, {\rm d}x &\overset{(1)}{=} \frac{\Gamma(a+1)}{\left ( i-1 \right )^{a+1}} \\
&\!\overset{a \in \mathbb{Z}}{=\! =\! =\!} \; \frac{a!}{\left ( i-1 \right )^{a+1}}\\
&=a! \left ( 1-i \right )^{-1-a}
\end{align*}

and similarly for the second.