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\title{Integral Fun Integral}
\author{Shreenabh Agrawal }
\date{\today}

\begin{document}

\maketitle

\section{Question}

$$I_{(\alpha, n)}=\int\limits_{0}^{\infty} \frac{\cos (\alpha x)}{x^{n}} \: d x$$

\section{Solution}
We Know,
$$\begin{aligned}
\int\limits_{0}^{\infty} f(t) \: g(t) \: d t=\int\limits_{0}^{\infty} \mathcal{L}\{f(t)\} \: \mathcal{L}^{-1}\{g(t)\} \: d t
\end{aligned}$$
Applying this,
$$\begin{aligned}
\therefore I_{(\alpha, n)} &=\int\limits_{0}^{\infty} \mathcal{L}\{\cos (\alpha x)\} \: \mathcal{L}^{-1}\left\{\frac{1}{x^{n}}\right\} \: d x \\
&=\int\limits_{0}^{\infty} \frac{s}{s^{2}+\alpha^{2}} \times \frac{s^{n-1}}{\Gamma(n)} \: d s \\
&=\frac{1}{\Gamma(n)} \int\limits_{0}^{\infty} \frac{s^{n}}{s^{2}+\alpha^{2}} \: d s
\end{aligned}$$
Taking Substitution,

$$\begin{aligned}
s&= \alpha \: \tan \theta \\
d s &=\alpha \: \sec ^{2} \theta \: d \theta
\end{aligned}$$

$$\begin{aligned} \Rightarrow \quad I_{(\alpha, n)} &=\frac{1}{\Gamma(n)} \int\limits_{0}^{\pi / 2} \frac{\alpha^{n} \tan ^{n} \theta}{\alpha^{2} \sec ^{2} \theta} \: \alpha \sec^{2} \theta \: d \theta \\ &=\frac{\alpha^{n-1}}{\Gamma(n)} \int\limits_{0}^{\pi / 2} \tan ^{n} \theta \: d \theta \end{aligned}$$

$$\boxed{I_{(\alpha, n)}=\frac{\pi \alpha^{n-1}}{2 \Gamma(n) \cos \left(\frac{\pi n}{2}\right)}}$$

\end{document}
