Finding the integral $\int_0^{\infty}\frac{\text{exp}\left(j \nu x\right)}{(x+1)^2}\,dx$

Question

I have this integral

$$\psi_X(j\nu)=E_X\left[e^{j\nu X}\right]=\int_0^{\infty}\frac{\text{exp}\left(j \nu x\right)}{(x+1)^2}\,dx$$

where $X\in[0,\infty)$ is a random variable with PDF $f_X(x)=(x+1)^{-2}$ and $j=\sqrt{-1}$. How can I evaluate this integral? I tried to use the Table of Integrals formulae, and found a close one, but the conditions to evaluate the integral aren't met. I tried integration by parts, but the same problem appears again, name, the real part of the exponential argument is 0, which doesn't have a solution in the Table of Integrals, as far as I know.

Thanks

What is $f$? If it is a contant, I'd say that it is not the best possible choice of letter... — ajotatxe, Sep 09 '16 at 15:14
What is the context? Share your thoughts on what you have tried. — StubbornAtom, Sep 09 '16 at 15:16
I changed the variable $f$ to $\nu$, and gave the context in the original post. The integral is basically the characteristic function of the random variable $X$, where the PDF of the random variable $X$ is $f_X(x)=1/(x+1)^2$. Thanks — EngDavid, Sep 09 '16 at 15:27

GEdgar · Accepted Answer · 2016-09-09T19:17:27.543

0

Maple says $$ \int_{0}^{\infty }\!{\frac {{{\rm e}^{i\nu\,x}}}{ \left( 1+x \right) ^ {2}}}\,{\rm d}x=1+{ {{{\rm e} ^{-i\nu}} i\,\nu}{\;{\rm Ei}_1 \left(-i\nu \right)}} $$ where the exponential integral $\mathrm{Ei}_1$ is $$ \mathrm{Ei}_1(z) := \int_1^\infty \frac{e^{-tz}}{t}\,dt $$

edited Sep 09 '16 at 19:17

answered Sep 09 '16 at 16:00

GEdgar

111,679

From Table of Integrals 7th edition eq 3.353.3 $\int_0^{\infty}e^{-px}/(a+x)^2 dx=pe^{ap}\text{Ei}(-ap)+(1/a)$. This gives the answer to my integral $-j\nu e^{-j\nu}\text{Ei}(j\nu)+1$. It's very close to yours, but not the same. However, the condition is that p>0. Can I use it in this case? – EngDavid Sep 09 '16 at 16:10
I left out an $i$, now corrected. You have $\mathrm{Ei}$ and Maple has $\mathrm{Ei}_1$, which should account for the differences in signs. – GEdgar Sep 09 '16 at 19:22
Thanks. I updated the original post. There is another part to it. – EngDavid Sep 10 '16 at 00:31

Michael_K · Answer 2 · 2016-09-09T20:30:15.763

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Using the substitution $t=x+1$ leads to

$$\int_1^{\infty}\frac{\text{exp}\left(j \nu (t-1)\right)}{t^2}\,dt = \text{exp}\left(-j \nu \right)\int_1^{\infty}\frac{\text{exp}\left(j \nu t\right)}{t^2}\,dt$$

Using integration by parts leads to

$$\text{exp}\left(-j \nu \right)\int_1^{\infty}\frac{\text{exp}\left(j \nu t\right)}{t^2}\,dt\\= 1 + \text{exp}\left(-j \nu \right)\int_1^{\infty}\frac{j\, \nu\,\text{exp}\left(j \nu t\right)}{t}\,dt = 1 + j \, \nu \, \text{exp}\left(-j \nu \right) \, E_1\left(-j\nu \right)$$

edited Sep 09 '16 at 20:30

answered Sep 09 '16 at 16:42

Michael_K

789

Thank you. This was a part of the problem. I will edit the original post to add the other part. – EngDavid Sep 09 '16 at 17:59
I think it's $1-j\nu e^{-j\nu}\text{Ei}(j\nu)$ isn't it? Because $\int_1^{\infty}\frac{e^{j\nu t}}{t}dt=E_1(-j\nu)=-\text{Ei}(j\nu)$. See https://en.wikipedia.org/wiki/Exponential_integral – EngDavid Sep 09 '16 at 18:13
You are of course right. I changed it in my post. – Michael_K Sep 09 '16 at 20:31
Thanks. I updated the original post. There is another part to it. – EngDavid Sep 10 '16 at 00:31

Finding the integral $\int_0^{\infty}\frac{\text{exp}\left(j \nu x\right)}{(x+1)^2}\,dx$

2 Answers2