# Calculating integral with exponential function and parameter

I want to calculate this integral but Mathematica does not solve it. Why? Can it be solved by Mathematica?

Integrate[Exp[-(Sqrt[1 + x^2] - 1)/t]*x^2, {x, 0, Infinity}]



I asked Mathematica to change variables to

 "u == Sqrt[1 + x^2] - 1"


and the result is Bessel function. but I'm not sure if

"IntegrateChangeVariables"


works properly.

This is the result:

ConditionalExpression[Exp^(1/t) t BesselK[2, 1/t], Re[t] > 0]

• Perhaps you have a typo... t-->x? Sep 3, 2023 at 14:46
• @OkkesDulgerci no, it's t
– lia
Sep 3, 2023 at 14:56
• Haha so this is a little awkward, I did exactly what you did for the add-on edit to your post in your question, and posted it at the exact same time you made the edit. I did some numerical confirmation of the result for explicit values of $t$ at the bottom of my answer though to confirm ConditionalExpression[Exp^(1/t) t BesselK[2, 1/t], Re[t] > 0] is in fact equal to the integral
– ydd
Sep 3, 2023 at 15:36
• Thank you so much for your help
– lia
Sep 3, 2023 at 15:43

You just need to change variables. Using $$u=\sqrt{x^2+1}$$ appears to work:

int = Exp[-(Sqrt[1 + x^2] - 1)/t]*x^2;

new = IntegrateChangeVariables[
Inactive[Integrate][int, {x, 0, Infinity}], u, u == Sqrt[1 + x^2]]

(*Inactive[Integrate][
E^((1 - u)/t) u Sqrt[-1 + u^2], {u, 1, \[Infinity]}]*)


Yields $$\int _1^{\infty }u \sqrt{u^2-1} e^{\frac{1-u}{t}}du$$

Activating the output:

Activate[new]
(*ConditionalExpression[E^(1/t) t BesselK[2, 1/t], Re[t] > 0]*)


So $$\int_0^{\infty } x^2 \exp \left(-\frac{\sqrt{x^2+1}-1}{t}\right) \, dx = e^{1/t} ~t~ K_2\left(\frac{1}{t}\right)$$

for $$\Re(t)>0$$, where $$K_n(t)$$ is the modified Bessel function of the second kind.

Looking experimentally with values of $$t$$, we see this is correct

Table[NIntegrate[int, {x, 0, Infinity}], {t, 5}]
(*{4.41677, 24.8963, 73.3853, 161.878, 302.373}*)

Table[(E^(1/t) t BesselK[2, 1/t]), {t, 5}] // N

(*{4.41677, 24.8963, 73.3853, 161.878, 302.373}*)

$$$$
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