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Good day! Is this integral tabular? I calculated it in MatLab and am now trying to write down an analytical expression. How can I get a result?

\begin{align} I &= \int\limits_{x=-\infty}^{-1} \frac{\mu \, dx}{2 \cdot (1+\mu^2 \cdot ((x-m) \cdot k)^2)^{3/2}} \\ &= \left[- \frac{\mu \cdot (m-x)}{2 k \mu \sqrt{(m-x)^2+1}} \right]_{x = -\infty}^{-1} \\ &= - \frac{\mu \cdot (m+1)}{2 k \mu \sqrt{(m+1)^2+1}} + \lim_{x\to-\infty} \, \frac{\mu \cdot (m-x)}{2 k \mu \sqrt{(m-x)^2+1}} \\ &= - \frac{\mu \cdot (m+1)}{2 k \mu \sqrt{(m+1)^2+1}} + \frac{\mu \cdot (m+\infty)}{2 k \mu \sqrt{(m+\infty)^2+1}} \\ &= \frac{1}{2 k} - \frac{\mu \cdot (m+1)}{2 k \mu \sqrt{(m+1)^2+1}}. \end{align}

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  • $\begingroup$ I guess you need to evaluate the limit as $x\to-\infty$. I get $-1/2k$. $\endgroup$
    – Patricio
    Nov 14, 2022 at 5:07
  • $\begingroup$ Thank you, I corrected the above, see if everything is written down correctly now? $\endgroup$
    – Антон
    Nov 14, 2022 at 5:20
  • $\begingroup$ I'm not understanding what you mean by "tabular". $\endgroup$
    – Teepeemm
    Nov 14, 2022 at 13:23

1 Answer 1

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The integral $$ I = \int\limits_{x=-\infty}^{-1} \frac{\mu \, dx}{2 \cdot (1+\mu^2 \, k^2 \, (x-m)^2)^{3/2}} $$ can be evaluated as follows. \begin{align} I &= \int\limits_{x=-\infty}^{-1} \frac{\mu \, dx}{2 \cdot (1+\mu^2 \cdot ((x-m) \cdot k)^2)^{3/2}} \\ &= \frac{\mu}{2} \, \int_{1}^{\infty} \frac{dt}{(1 + \mu^2 \, k^2 \, (m+t)^2)^{3/2}} \hspace{5mm} \to x = -t \\ &= \frac{\mu}{2} \, \left[ \frac{m+t}{\sqrt{\mu^2 \, k^2 \, (m+t)^2 + 1}} \right]_{1}^{\infty} \\ &= \frac{\mu}{2} \, \left[ \lim_{t \to \infty} \frac{1}{\sqrt{\mu^2 \, k^2 + \frac{1}{(m+t)^2}}} - \frac{m+1}{\sqrt{\mu^2 \, k^2 \, (m+1)^2 + 1}} \right] \\ &= \frac{\mu}{2} \, \left[ \frac{1}{\mu \, k} - \frac{m+1}{\sqrt{\mu^2 \, k^2 \, (m+1)^2 + 1}} \right] \\ &= \frac{1}{2 \, k} - \frac{\mu \, (m+1)}{2 \, \sqrt{\mu^2 \, k^2 \, (m+1)^2 + 1}} \end{align}

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  • $\begingroup$ Thank you so much, he thinks everything is right! Tell me, in the second line you brought the integral to a tabular form? How can I find it in the table of integrals? $\endgroup$
    – Антон
    Nov 14, 2022 at 6:44
  • $\begingroup$ @Антон There may be integral tables that have some form of related integrals, but in general $$ \frac{d}{dx} \, \frac{1}{\sqrt{f(x)}} = - \frac{f^{'}(x)}{( f(x) )^{3/2}} $$. From here it is a matter of finding the appropriate $f(x)$ that matches the integral. $\endgroup$
    – Leucippus
    Nov 14, 2022 at 6:55

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