# Is there a closed form solution available for following integral

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?

• Where did you catch this monster ? Sep 10 '17 at 13:08
• have you asked Wolfram alpha? Sep 10 '17 at 13:08
• U can try using Complex Analysis and solve it by Contour integration Sep 10 '17 at 13:10
• That's almost strange of you to ask @Dr.SonnhardGraubner Sep 10 '17 at 13:12
• @ClaudeLeibovici I am trying to solve a research problem related to exponential random variables Sep 10 '17 at 13:13

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 $$\frac {a\,e^{-dx}}{b+c}\Big[1-\frac{2 b}{(b+c)} \frac 1{x^{1/2}}+\frac{a b (3 b-c)}{(b+c)^3} \frac 1x+\cdots\Big]$$
• $O(1/x)$ doesn't guarantee convergence. Feb 6 at 16:13
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 $\surd\pi$.