2
$\begingroup$

I need to solve following integral $$\int_0^{\infty}\frac{ax}{b(\sqrt{x}+1)^2+cx}e^{-dx}dx$$ where $a>0,b>0,c>0,d>0$. Any help in this regard will be appreciated.

My Attempt: I have an idea to solve the above integral. For small values of $x$ I represent $e^{-d*x}$ as a series and for large values of $x$ I represent $\frac{ax}{b(\sqrt{x}+1)^2+cx}$ as a series. In this way I can get an answer numerically. Is it a good idea? Further, how should I decide about the critical value of $x$ where the switch occurs?

$\endgroup$
  • 1
    $\begingroup$ Where did you catch this monster ? $\endgroup$ – Claude Leibovici Sep 10 '17 at 13:08
  • 1
    $\begingroup$ have you asked Wolfram alpha? $\endgroup$ – Dr. Sonnhard Graubner Sep 10 '17 at 13:08
  • $\begingroup$ U can try using Complex Analysis and solve it by Contour integration $\endgroup$ – ys wong Sep 10 '17 at 13:10
  • 1
    $\begingroup$ That's almost strange of you to ask @Dr.SonnhardGraubner $\endgroup$ – Simply Beautiful Art Sep 10 '17 at 13:12
  • $\begingroup$ @ClaudeLeibovici I am trying to solve a research problem related to exponential random variables $\endgroup$ – Frank Moses Sep 10 '17 at 13:13
0
$\begingroup$

The only thing I could say is that there will not be any problem to compute since, with $a>0$ ,$b>0$, $c>0$, $d>0$ :

  • around $x=0$, the integrand is $$\frac{a x}{b}-\frac{2 a x^{3/2}}{b}+O\left(x^2\right)$$
  • for large values of $x$, the integrand is $$\frac {a\,e^{-dx}}{b+c}+O\left(\frac 1x\right)$$
$\endgroup$
0
$\begingroup$

The pre-factor has a taylor series in $\sqrt x$ around $x=0$, and interchange of intergration and summation leads to terms of the form $\int_0^\infty x^{\nu-1} e^{-d x}dx=d^{-\nu}\Gamma(\nu)$. The $\Gamma$-values are either factorials or multiples of $\surd\pi$.

$\endgroup$
  • $\begingroup$ can you please write the Taylor series that you mentioned in your comment $\endgroup$ – Frank Moses Sep 10 '17 at 16:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.