# Definite integral $\int_{0}^{+\infty} \exp(-\sqrt{x^2+bx+c}) dx$

I would like to calculate the definite integral $$I_1 = \int_{0}^{\infty} \exp(-\sqrt{x^2+bx+c})dx$$ where $$b$$ and $$c$$ are reals. I feel that the solution, if there is one, has something to do with Bessel functions. I've first tried the following substitution $$t = \sqrt{x^2+bx+c}$$ which gave me $$I_1 = \int_{0}^{\infty} \exp(-t) \frac{2t}{\sqrt{b^2-4(c-t^2)}}dt$$ but I don't know where to go after that step. It doesn't look like any of the famous integrals of exponential functions and I haven't been able to integrate it by parts.

EDIT : Accelerator pointed out to me that the integration bounds shouldn't be $$[0,\infty[$$ (there isn't always a solution for $$t=0$$ if $$b$$ and $$c$$ are reals), which is absolutely right. I had overlooked this point and will try to modify my question in line with this remark.

I also tried the substitution $$t = x+b/2$$ to shift the parabola and obtain $$I_1 = \int_{b/2}^{\infty} \exp(-\sqrt{t^2+z})dt,$$ with $$z=c-b^2/4$$, which is the same integral as in this question but with a non-zero lower bound.

I also know that my problem can be formulated with another integral, $$I_2 = \int_{\theta_0}^{\pi/2} \exp\left(-\frac{k}{cos(\theta)}\right )d\theta,$$ where $$k$$ is a real number, but I don't know if this can be helpful as we again have a non-zero lower bound.

Ideally I would like to obtain the indefinite integral but the integral between zero and infinity would already be really useful.

• One incomplete idea: Using Euler's first substitution, we could define $$t +x = \sqrt{x^2 + bx + c}$$ and hence have $$x = \frac{c-t^2}{2t-b}$$ and $$dx = - 2 \cdot \frac{t^2 - bt + c}{(2t-b)^2} \, dt$$ Note that, when $x=0$, then $t = \sqrt c$, and as $x \to \infty$, $t\to b/2$. [cont.] Commented Jun 8, 2023 at 22:50
• This gives us the integral $$- 2 \int_{\sqrt c}^{b/2} \exp \left( -\frac{t^2 -bt+c}{2t-b} \right) \frac{t^2 - bt + c}{(2t-b)^2} \, dt$$ I wonder if something can be done from here, considering the repeated structure. Note, too, that $$\frac{d}{dt} (t^2 - bt + c) = 2t-b$$ Commented Jun 8, 2023 at 22:50
• Did you phrase the problem correctly? There isn't an $a$ anywhere and the integrand doesn't seem to be defined for all real numbers $b,c$ and $0 < x < \infty$, unless it works for complex numbers. Commented Jun 8, 2023 at 23:15
• @Accelerator you're right. I started to write my question for the general case $-\sqrt{ax^2+bx+c}$ but in my case $a=1$ so I removed this constant from the equations but I missed one. And concerning the integrand you're also right, in my problem $c$ is always a positive real and $b$ can be a negative or positive real so the lower bound of integration should not be always $0$. I'll remove the $a$ constant first and modify my bounds later as I don't have a lot of time right now. Sorry for the confusing question, I checked before posting it but it was late and I was very tired. Commented Jun 9, 2023 at 8:06
• That's fine. If you want to redo everything, it's better to ask about the integral with $a,b,c$ and a different domain of integration in a separate post to avoid getting flagged and having a moderator step in. You've already received an answer concerning your initial upload w/ no edits. Since square roots aren't defined for negative numbers, you can probably solve $0 < ax^2 + bx + c$ to see what values of $a,b,c$ make the inside of the square root negative. There's also the issue of choosing the domain of integration, if your goal is to construct a random integral that's possible to solve. Commented Jun 9, 2023 at 8:36

One of my past PhD students faced almost he sam problem for the case whare $$x^2+bx+c$$ is always positive and $$\left(c-\frac{b^2}{4}\right)$$ being "small".
Completing the square and changing notations, the integrand write $$e^{-\sqrt{(x+\alpha )^2+\beta}}$$
What she did was to expand it as a series around $$\beta=0$$ $$e^{-\sqrt{(x+\alpha )^2+\beta}}=\sum_{n=0}^\infty A_n\,\beta^n$$ with $$A_0=e^{-(x+\alpha)} \qquad \qquad A_1=-\frac{e^{-(x+\alpha)}}{2 (x+\alpha )}$$ $$A_n=-\frac{2 (n-1) (2 n-3) A_{n-1}-A_{n-2} } {4 n(n-1)(x+\alpha )^2 }$$
In other words, $$e^{-\sqrt{(x+\alpha )^2+\beta}}=e^{-(x+\alpha )} \Big[\cdots\Big]$$ where $$\Big[\cdots\Big]=1-\frac{1}{2 (x+\alpha )}\beta+\left(\frac{1}{8 (x+\alpha )^2}+\frac{1}{8 (x+\alpha )^3}\right)\beta ^2 -$$ $$\left(\frac{1}{48 (x+\alpha )^3}+\frac{1}{16 (x+\alpha )^4}+\frac{1}{16 (x+\alpha )^5}\right)\beta ^3 +O\left(\beta ^4\right)$$ leading to simple integrals since $$I_n=\int_0^\infty \frac{e^{-(x+\alpha)}} {(x+\alpha)^{n}}=\Gamma (1-n,\alpha )$$
For a quick test $$(\alpha=2,\beta=3)$$, using the very truncated series given above leads to $$0.0798$$ to be compared to the "exact" $$0.0832$$. Using twice more terms leads to $$0.0836$$