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First of all i know that there were a similar question: Evaluating $\int_0^\infty\frac{\sin(x)}{x^2+1}\, dx$

However i am going to ask for something different

Let there be two integrals

$I_{1}(\alpha) = \int_{0}^{\infty}\frac{\cos(\alpha x)}{(b+1)x^2+2ix+(b-1)}dx$

$I_{2}(\alpha)= \int_{0}^{\infty}\frac{\sin(\alpha x)}{(b+1)x^2+2ix+(b-1)}dx$

Where $b$ is a complex number such $|b|<1$

Find $I_{1},I_{2}$

Here is my attempt

I took function $f(z)=\frac{e^{i\alpha z}}{(b+1)z^2+2iz+(b-1)}$

It has poles at $z=-i$ and $z=i\frac{b-1}{b+1}$

And i took a part of a circle with two radius as a contour. However i think it is not a good way to solve it.

mkultra
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1 Answers1

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Probably the simplest way to evaluate the integrals is shown in the link you have given. I will assume $\alpha$ to be a positive real number.

Define: $$\begin{align} I(\alpha)=\int_{0}^{\infty}\frac{e^{-\alpha x}dx}{(b+1)x^2+2ix+(b-1)} &=\frac i{2b}\int_{0}^{\infty}e^{-\alpha x}\left[\frac1{x+i}-\frac1{x+i\frac{1-b}{1+b}}\right]dx. \end{align}$$ Now $$ \int_{0}^{\infty}\frac{e^{-\alpha x}}{x+i}dx =e^{i\alpha}\int_{0}^{\infty}\frac{e^{-\alpha (x+i)}}{x+i}dx \stackrel{t=\alpha(x+i)}=e^{i\alpha}\int_{i\alpha}^{\infty}\frac{e^{-t}}{t}dt =e^{i\alpha}E_1(i\alpha), $$ where $E_1(z)$ is the exponential integral. After similar calculation of the second term one obtains: $$ I(\alpha)=\frac i{2b}\left[e^{i\alpha} E_1(i\alpha)-e^{i\beta} E_1(i\beta)\right], \quad\text{with}\quad \beta=\alpha\frac{1-b}{1+b}. $$

Finally $$ I_1(\alpha)=\frac{I(-i\alpha)+I(i\alpha)}2;\quad I_2(\alpha)=\frac{I(-i\alpha)-I(i\alpha)}{2i}. $$

user
  • 26,272