3
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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.] $\endgroup$ Commented Jun 8, 2023 at 22:50
  • $\begingroup$ 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 $$ \frac{d}{dt} (t^2 - bt + c) = 2t-b $$ $\endgroup$ Commented Jun 8, 2023 at 22:50
  • $\begingroup$ 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. $\endgroup$ Commented Jun 8, 2023 at 23:15
  • $\begingroup$ @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. $\endgroup$ Commented Jun 9, 2023 at 8:06
  • $\begingroup$ 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. $\endgroup$ Commented Jun 9, 2023 at 8:36

2 Answers 2

Reset to default
2
$\begingroup$

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 $0.0836$

$\endgroup$
1
$\begingroup$

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

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:

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

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.