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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.

Eliam
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  • 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.] – PrincessEev Jun 08 '23 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 $$

     – PrincessEev Jun 08 '23 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. – Accelerator Jun 08 '23 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. – Eliam Jun 09 '23 at 08:06
  • 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. – Accelerator Jun 09 '23 at 08:36

2 Answers2

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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 $0.0836$

Claude Leibovici
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$$\int_{0}^{\infty} e^{-\sqrt{x^2+bx+c}}dx$$

If $\left(x+q\right)^2=x^2+bx+c,\quad c=q^2, \quad b=2q$:

$$\int_{0}^{\infty} e^{-\sqrt{x^2+2qx+q^2}}dx$$

$$u=\left(x+q\right)$$

$$\boxed{\int_{\sqrt{c}}^{\infty} e^{-u}du = e^{-\sqrt{c}}}$$

General case $\left(x+\frac{b}{2}\right)^2+\left(\frac{4c-b^2}{4}\right)$:

$$\int_{0}^{\infty}e^{-\sqrt{\left(x+\frac{b}{2}\right)^2+\left(\frac{4c-b^2}{4}\right)}}dx$$ $$u= x+\frac{b}{2},\quad \phi = \frac{4c-b^2}{4}$$ $$\int_{\frac{b}{2}}^{\infty} e^{-\sqrt{u^2 + \phi}}du$$

The general result is in the form of the modified Bessel functions:

$$\boxed{\int_{0}^{\infty} e^{-\sqrt{x^2+bx+c}}dx=\frac{1}{2}e^{-\frac{b}{2}}bK_0\left(\sqrt{\frac{4c-b^2}{4}}\right) +e^{-\frac{b}{2}}\sqrt{\frac{4c-b^2}{4}}K_1\left(\sqrt{\frac{4c-b^2}{4}}\right)}$$

Daniel K
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