3

$$ \int_{-\infty}^{-1} e^{ikx} \left( \frac{-A}{-x-1+\sqrt{x^2-1}} \right)dx = \frac{A}{2}\int_1^\infty e^{-ikx} \left( 1 - \sqrt{\frac{x+1}{x-1}} \right)dx. $$ Hello, thank you very much for this website, I want to know how is it possible to express the integral in the left side as the integral in the right side? I have tried rationalizing the denominator in the left side but nothing happens. (I know that the integral in the right side can be expressed in terms of the modified bessel functions of the second kind of order zero and order one.) However,I don't understand how can the integral be expressed from the left side to the right side

Thank you very much

Siminore
  • 35,136
copets
  • 31
  • It does not answer your question, but maybe you can find this guide useful, if you want to learn how to type in MathJax. – Bman72 Oct 28 '14 at 12:05
  • $-x-1=-(x+1)=-(\sqrt{x+1})^2$ so you can factorize the bottom line and go from there. – Paul Oct 28 '14 at 12:11
  • Infact, considering the range of integration, better to change variables first so that the integral goes 1 to infinity. Now take out an $\sqrt{x-1}$ factor on the bottom line and use the difference of 2 squares to finish. – Paul Oct 28 '14 at 12:37
  • Thank you, I can follow the factorization, but how to change the variables to make the integral between 1 and infinity? – copets Oct 28 '14 at 13:05

1 Answers1

0

Set $x\equiv-x$ then your integral can be rewritten as \begin{align}\int^{-1}_{-\infty}e^{ikx}\frac{-A}{-x-1+\sqrt{x^2-1}}\,dx&=-\int^{1}_{\infty}e^{-ikx}\frac{-A}{x-1+\sqrt{x^2-1}}\,dx\\&=\int^{\infty}_{1}e^{-ikx}\frac{-A}{x-1+\sqrt{x^2-1}}\,dx\\ &=\int^{\infty}_{1}e^{-ikx}\frac{-A(x-1-\sqrt{x^2-1})}{(x-1)^2-(x^2-1)}\,dx\\ &=\int^{\infty}_{1}e^{-ikx}\frac{-A(x-1-\sqrt{x^2-1})}{2-2x}\,dx\\ &=\int^{\infty}_{1}e^{-ikx}\frac{A(x-1-\sqrt{x^2-1})}{2(x-1)}\,dx\\ &=\frac{A}{2}\int^{\infty}_{1}e^{-ikx}\frac{(x-1-\sqrt{x^2-1})}{x-1}\\& =\frac{A}{2}\int^{\infty}_{1}e^{-ikx}(1-\sqrt\frac{x+1}{x-1})\,dx\end{align}

Arian
  • 6,277
  • Thank you very much, how you flip the sign in the first step? did you multiply the variable of integration x by a minus sign such that the sign of the limits of integration change? – copets Oct 28 '14 at 13:29
  • @copets: yes basically I made the substitution $x\to-x$ and the sign of the limits of integration change. – Arian Oct 28 '14 at 13:41
  • @copets:you're welcome! :) – Arian Oct 28 '14 at 13:50