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$$$$ The following integral is a divergent integral:

$$\int_{0}^{\infty} \frac{\mathrm{e}^{- a\, x}\, \sin\!\left(x\right)}{x^5} \,d x $$

However, the following solution is provided (though divergent solution) in the book of "Theory of elasticity of microheterogeneous media", by Shermergor (in Russian):

$$\frac{25\, a}{72} + \frac{\mathrm{\delta}\, \left(3\, a^2 - 1\right)}{6} - \frac{a\, {\mathrm{\delta}}^2}{2} - \frac{25\, a^3}{72} + \frac{\arctan\!\left(\frac{1}{a}\right)\, \left(a^4 - 6\, a^2 + 1\right)}{24} + \frac{{\mathrm{\delta}}^3}{3} + \frac{a\, \mathrm{log}\!\left(a^2 + 1\right)\, \left(a^2 - 1\right)}{12}$$

Where $\mathrm{\delta}$ is an infinite constant. Can anybody help me to understand how this solution is obtained? Is there any logical hierarchy to obtain analogous solutions for the general case of:

$$\int_{0}^{\infty} \frac{\mathrm{e}^{- a\, x}\, \sin\!\left(x\right)}{x^n} \,d x $$

Thank you so much for the time,

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  • $\begingroup$ I believe you didn't copy the formula from the book correctly. There should be an $I_c$ term, where $I_c = \int_{1/\delta}^\infty x^{-1} e^{-a x} \cos x \,dx$. $\endgroup$ – Maxim Oct 17 '18 at 0:09
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Well, we are looking at:

$$\mathcal{I}_{\space\text{n}}\left(\alpha\right):=\int_0^\infty\frac{\exp\left(-\alpha\cdot x\right)\cdot\sin\left(x\right)}{x^\text{n}}\space\text{d}x\tag1$$

Using the evaluating integrals over the positive real axis property of the Laplace transform we can write:

$$\mathcal{I}_{\space\text{n}}\left(\alpha\right)=\int_0^\infty\mathscr{L}_x\left[\exp\left(-\alpha\cdot x\right)\cdot\sin\left(x\right)\right]_{\left(\text{s}\right)}\cdot\mathscr{L}_x^{-1}\left[\frac{1}{x^\text{n}}\right]_{\left(\text{s}\right)}\space\text{d}\text{s}\tag2$$

Using the table of selected Laplace transforms, we can write:

  • $$\mathscr{L}_x\left[\exp\left(-\alpha\cdot x\right)\cdot\sin\left(x\right)\right]_{\left(\text{s}\right)}=\frac{1}{1+\left(\alpha+\text{s}\right)^2}\tag3$$
  • $$\mathscr{L}_x^{-1}\left[\frac{1}{x^\text{n}}\right]_{\left(\text{s}\right)}=\frac{\text{s}^{\text{n}-1}}{\Gamma\left(\text{n}\right)}\tag4$$

So, we get:

$$\mathcal{I}_{\space\text{n}}\left(\alpha\right)=\int_0^\infty\frac{1}{1+\left(\alpha+\text{s}\right)^2}\cdot\frac{\text{s}^{\text{n}-1}}{\Gamma\left(\text{n}\right)}\space\text{d}\text{s}=\frac{1}{\Gamma\left(\text{n}\right)}\cdot\int_0^\infty\frac{\text{s}^{\text{n}-1}}{1+\left(\alpha+\text{s}\right)^2}\space\text{d}\text{s}\tag5$$

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