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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 $$f_n(\alpha,\beta)=\int_0^{\infty}\mathrm{e}^{-x^n}\sin(\alpha x)\cos(\beta x)\,dx$$

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    $\begingroup$ If $n > 1$, expanding $e^{i (\pm \alpha \pm \beta) x}$ into series gives a sum of four Fox-Wright functions. If additionally $n \in \mathbb N$, the result can be reduced to a sum of generalized hypergeometric functions (essentially by applying the Gauss-Legendre multiplication theorem). $\endgroup$
    – Maxim
    Jun 5 '20 at 6:10
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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:

  • $$\text{I}_1=\int_0^\infty\mathcal{L}_x\left[\sin\left(\left(\alpha-\beta\right)x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[x^\text{kn}\right]_{\left(\text{s}\right)}\space\text{ds}\tag6$$
  • $$\text{I}_2=\int_0^\infty\mathcal{L}_x\left[\sin\left(\left(\alpha+\beta\right)x\right)\right]_{\left(\text{s}\right)}\cdot\mathcal{L}_x^{-1}\left[x^\text{kn}\right]_{\left(\text{s}\right)}\space\text{ds}\tag7$$

And using the table of selected Laplace transforms, we have:

  • $$\mathcal{L}_x\left[\sin\left(\left(\alpha-\beta\right)x\right)\right]_{\left(\text{s}\right)}=\frac{\alpha-\beta}{\left(\alpha-\beta\right)^2+\text{s}^2}\tag8$$
  • $$\mathcal{L}_x\left[\sin\left(\left(\alpha+\beta\right)x\right)\right]_{\left(\text{s}\right)}=\frac{\alpha+\beta}{\left(\alpha+\beta\right)^2+\text{s}^2}\tag9$$
  • $$\mathcal{L}_x^{-1}\left[x^\text{kn}\right]_{\left(\text{s}\right)}=\frac{1}{\text{s}^{1+\text{kn}}}\cdot\frac{1}{\Gamma\left(-\text{kn}\right)}\tag{10}$$

In order to finish you can use this.

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  • $\begingroup$ @JanEarland, it’s you again, and can you please tell me how to finish it because when I went to where you have written it I couldn’t understand somethings so please can you finish this, it will also help me understand my last question $\endgroup$
    – Asv
    Jun 4 '20 at 19:22

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