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I was working in a problem in physics, which gave me this integral, and I need solution:

$$

\int_{-\infty}^\infty \exp{\left(-\sqrt{x^2 + a^2}\right)}dx

$$

The problem is, I have no clue how to start. I tried several substitution of variables, none of them successful. If any of you point out some hints of clues, or even how to get to solution, or resources/references/recomendations, it would be very appreciated.

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Put $x=a\sinh(t)$, then we obtain after using parity of the original integrand

$$

I(a)=2a\int_{0}^{\infty}e^{-a\cosh(t)}\cosh(t)dt

$$

This equals 10.32.9 here and therefore

$$

I(a)=2aK_1(a)

$$

where $K_{\nu}(z)$ denotes a modified Bessel-function of the second type

**Edit**

To make this answer a little bit more self contained, i proof the stated integral representation of the Bessel function

We need to show that

$$

F_{\nu}(a)=\int_{0}^{\infty}e^{-a\cosh(t)}\cosh(\nu t)dt=K_{\nu}(a) \quad (\star)

$$

To do so let’s have a look at $F^{”}_{\nu}(a)$, where the $’$ denotes a derivative w.r.t. to $a$. We get

$$

F^{”}_{\nu}(a)=\int_{0}^{\infty}e^{-a\cosh(t)}\cosh^2( t)\cosh(\nu t)dt=\\

\int_{0}^{\infty}e^{-a\cosh(t)}(1+\sinh^2( t))\cosh(\nu t)dt=\\

F_{\nu}(a)+\int_{0}^{\infty}e^{-a\cosh(t)}\sinh^2( t)\cosh(\nu t)dt

$$

Here we used $(\star)$ in the last line.

Next let’s integrate by parts

$$

F^{”}_{\nu}(a)=F_{\nu}(a)+\frac{1}{a}\int_{0}^{\infty}e^{-a\cosh(t)}(\cosh( t)\cosh(\nu t)+\nu\sinh( t)\sinh(\nu t))dt=\\

F_{\nu}(a)-\frac{1}{a}F^{‘}_{\nu}(a)+\frac{\nu}{a}\int_{0}^{\infty}e^{-a\cosh(t)}\sinh( t)\sinh(\nu t)dt

$$

Here the first derivative of $(\star)$ was used.

Now another integration by parts reveals

$$

F^{”}_{\nu}(a)=F_{\nu}(a)-\frac{1}{a}F^{‘}_{\nu}(a)+\frac{\nu^2}{a^2}\int_{0}^{\infty}e^{-a\cosh(t)}\cosh(\nu t)dt

$$

and therefore

$$

F^{”}_{\nu}(a)=F_{\nu}(a)-\frac{1}{a}F^{‘}_{\nu}(a)+\frac{\nu^2}{a^2}F_{\nu}(a)

$$

which means that $F^{”}_{\nu}(a)$ fullfils the modified Bessel equation and $(\star)$ is indeed true.

To justify that we really have found a rep for $K_{\nu}(a)$ we note that $F_{\nu}(a)$ decays to zero at $a\rightarrow \infty$ in contrast to the other solution of the modified Bessel equation $I_{\nu}(a)$ which is growing exponentially for large arguments.

**QED**

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