I was just looking at the following function but couldn’t understand how to form a way of integrating this and how does it depend upon $n, \alpha$ and $\beta$ can anyone please help
$$f_n(\alpha,\beta)=\int_0^{\infty}\mathrm{e}^{-x^n}\sin(\alpha x)\cos(\beta x)\,dx$$
Evaluate $f_n(\alpha,\beta)=\int_0^{\infty}\mathrm{e}^{-x^n}\sin(\alpha x)\cos(\beta x)\,dx$
definite integralsintegration
Related Solutions
We make use of the identity
$$ \sum_{n=-\infty}^{\infty} \frac{1}{a^{2} - (x + n\pi)^{2}} = \frac{\cot(x+a) - \cot(x-a)}{2a}, \quad a > 0 \text{ and } x \in \Bbb{R}. $$
Then for $\alpha, \beta > 0$ it follows that
\begin{align*} I := \mathrm{PV}\int_{0}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}} &= \frac{1}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}} \, dx \\ &= \frac{\alpha}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - x^{2}} \, dx \\ &= \frac{\alpha}{2} \sum_{n=-\infty}^{\infty} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \, dx \\ &= \frac{\alpha}{2} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \right) \log\cos^{2}x \, dx \\ &= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx, \end{align*}
where interchanging the order of integration and summation is justified by Tonelli's theorem applied to the summation over large indices $n$. Then
\begin{align*} I &= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx \\ &= \frac{1}{2\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\left|2\cos x\right| \, dx \tag{1} \end{align*}
Here, we exploited the following identity to derive (1).
$$ \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot(x+a) \, dx = 0 \quad \forall a \in \Bbb{R}. $$
Now with the substitution $z = e^{2ix}$ and $\omega = e^{2i\alpha\beta}$, it follows that
\begin{align*} I &= \frac{1}{2\beta} \Re \mathrm{PV} \int_{|z|=1} \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \log(1 + z) \, \frac{dz}{z}. \tag{2} \end{align*}
Now consider the following unit circular contour $C$ with two $\epsilon$-indents $\gamma_{\omega,\epsilon}$ and $\gamma_{\bar{\omega},\epsilon}$.
Then the integrand of (2)
$$ f(z) = \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \frac{\log(1 + z)}{z} $$
is holomorphic inside $C$ (since the only possible singularity at $z = 0$ is removable) and has only logarithmic singularity at $z = -1$. So we have
$$ \oint_{C} f(z) \, dz = 0. $$
This shows that
\begin{align*} I &= \frac{1}{2\beta} \Re \lim_{\epsilon \downarrow 0} \left( \int_{-\gamma_{\omega,\epsilon}} f(z) \, dz + \int_{-\gamma_{\bar{\omega},\epsilon}} f(z) \, dz \right) \\ &= \frac{1}{2\beta} \Re \left( \pi i \mathrm{Res}_{z=\omega} f(z) + \pi i \mathrm{Res}_{z=\bar{\omega}} f(z) \right) \\ &= \frac{1}{2\beta} \Re \left( - \pi i \log(1 + \omega) + \pi i \log(1 + \bar{\omega}) \right) \\ &= \frac{\pi}{\beta} \arg(1 + \omega) = \frac{\pi}{\beta} \arctan(\tan (\alpha \beta)). \end{align*}
In particular, if $\alpha\beta < \frac{\pi}{2}$ then we have
$$ I = \pi \alpha. $$
But due to the periodicity of $\arg$ function, this function draws a scaled saw-tooth function for $\alpha > 0$. Of course, $I$ is an even function of both $\alpha$ and $\beta$, so the final result is obtained by even extension of this saw-tooth function.
If you don't mind, I would like to present an alternative approach that makes use of the fact that $$\int^\infty_0\frac{x^{p-1}}{1+x}dx=\frac{\pi}{\sin{p\pi}}$$ Simply factorise the denominator and decompose the integrand into partial fractions. \begin{align} \int^\infty_0\frac{x^a}{x^2+2(\cos{b})x+1}dx &=\int^\infty_0\frac{x^a}{(x+e^{ib})(x+e^{-ib})}dx\\ &=\frac{1}{-e^{ib}+e^{-ib}}\int^\infty_0\frac{x^a}{e^{ib}+x}dx+\frac{1}{-e^{-ib}+e^{ib}}\int^\infty_0\frac{x^a}{e^{-ib}+x}dx\\ &=\frac{1}{-2i\sin{b}}\int^\infty_0\frac{(e^{ib}u)^a}{1+u}du+\frac{1}{2i\sin{b}}\int^\infty_0\frac{(e^{-ib}u)^a}{1+u}du\\ &=\frac{e^{iab}}{-2i\sin{b}}\frac{\pi}{\sin(\pi a+\pi)}+\frac{e^{-iab}}{2i\sin{b}}\frac{\pi}{\sin(\pi a+\pi)}\\ &=\frac{\pi}{\sin \pi a\sin{b}}\left(\frac{e^{iab}-e^{-iab}}{2i}\right)\\ &=\frac{\pi\sin{ab}}{\sin{\pi a}\sin{b}} \end{align}
Best Answer
Well, we have the following integral:
$$\mathcal{I}_\text{n}\left(\alpha,\beta\right):=\int_0^\infty\exp\left(-x^\text{n}\right)\sin\left(\alpha x\right)\cos\left(\beta x\right)\space\text{d}x\tag1$$
Using the definition of the Exponential function:
$$\exp(x)=\sum_{\text{k}\ge0}\frac{x^\text{k}}{\text{k}!}\tag2$$
So, we can write:
$$\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}}{\text{k}!}\int_0^\infty x^\text{kn}\sin\left(\alpha x\right)\cos\left(\beta x\right)\space\text{d}x\tag3$$
Now, we also know that:
$$\sin\left(\alpha x\right)\cos\left(\beta x\right)=\frac{\sin\left(\left(\alpha-\beta\right)x\right)+\sin\left(\left(\alpha+\beta\right)x\right)}{2}\tag4$$
So:
$$\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\sum_{\text{k}\ge0}\frac{\left(-1\right)^\text{k}}{2\left(\text{k}!\right)}\left\{\underbrace{\int_0^\infty x^\text{kn}\sin\left(\left(\alpha-\beta\right)x\right)\space\text{d}x}_{\text{I}_1}+\underbrace{\int_0^\infty x^\text{kn}\sin\left(\left(\alpha+\beta\right)x\right)\space\text{d}x}_{\text{I}_2}\right\}\tag5$$
Now, we can use the 'evaluating integrals over the positive real axis' property of the Laplace transform in order to write:
And using the table of selected Laplace transforms, we have: