1. ## Tricky Exponential Integral

I'm asked to evaluate the following integral:

I know the classic gaussian result, but I cannot seem to use this to help. I'm given a hint to try the substitution: $y&space;=&space;\frac{1}{2}(a^{^{1/2}}x&space;-&space;\frac{b^{^{1/2}}}{x})$ , but I cannot make this work.

2. Here's my stab at it. Denote the integral by $I$. Because of the symmetry this integral is

$I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$

Under the change of variable $x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.

Now replace the dummy variable with $x$, so

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\dfrac{dx}{x^2},\;\;\;(2)$.

Now add $\sqrt{a} \;(1) + \sqrt{a} \;(2)$ giving

$2 \sqrt{a} I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;\left(\sqrt{a}+ \dfrac{\sqrt{b}}{x^2}\right)\;dx\;\;\;(3)$

Now introduce your hint (sort of)

$y = \sqrt{a} x - \dfrac{\sqrt{b}}{x}$

so

$dy = \left(\sqrt{a} + \dfrac{\sqrt{b}}{x^2}\right)\,dx$.

Note that $x \to 0$ gives $y \to - \infty$ and $x \to \infty$ gives $y \to \infty$ so with that and noting that

$y^2 = ax^2 - 2 \sqrt{ab} + \dfrac{b}{x^2}$ gives $ax^2 + \dfrac{b}{x^2} = y^2 + 2 \sqrt{ab}$.

Thus, (3) becomes

$2 \sqrt{a} I = 2\int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(y^2 + 2 \sqrt{ab}\right)}\;\;dy = 2e^{-\sqrt{ab}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2}\;\;dy$

and solving for $I$ gives

$I = \sqrt{\dfrac{2\pi}{a}} e^{-\sqrt{ab}}$.

Whew! I think I need to lie down.

3. Wow! Thanks.
Originally Posted by Danny
Here's my stab at it. Denote the integral by $I$. Because of the symmetry this integral is

$I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$

Under the change of variable $x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.

Now replace the dummy variable with $x$, so

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\dfrac{dx}{x^2},\;\;\;(2)$.

Now add $\sqrt{a} \;(1) + \sqrt{a} \;(2)$ giving

$2 \sqrt{a} I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;\left(\sqrt{a}+ \dfrac{\sqrt{b}}{x^2}\right)\;dx\;\;\;(3)$

Now introduce your hint (sort of)

$y = \sqrt{a} x - \dfrac{\sqrt{b}}{x}$

so

$dy = \left(\sqrt{a} + \dfrac{\sqrt{b}}{x^2}\right)\,dx$.

Note that $x \to 0$ gives $y \to - \infty$ and $x \to \infty$ gives $y \to \infty$ so with that and noting that

$y^2 = ax^2 - 2 \sqrt{ab} + \dfrac{b}{x^2}$ gives $ax^2 + \dfrac{b}{x^2} = y^2 + 2 \sqrt{ab}$.

Thus, (3) becomes

$2 \sqrt{a} I = 2\int_{-\infty}^{\infty} e^{-\frac{1}{2}\left(y^2 + 2 \sqrt{ab}\right)}\;\;dy = 2e^{-\sqrt{ab}}\int_{-\infty}^{\infty} e^{-\frac{1}{2}y^2}\;\;dy$

and solving for $I$ gives

$I = \sqrt{\dfrac{2\pi}{a}} e^{-\sqrt{ab}}$.

Whew! I think I need to lie down.

4. Originally Posted by Danny
Here's my stab at it. Denote the integral by $I$. Because of the symmetry this integral is

$I = 2\int_0^{\infty} e^{-\frac{1}{2}\left(ax^2 + \frac{b}{x^2}\right)}\;dx.\;\;\;(1)$

Under the change of variable $x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$, the integral becomes

$I =2\sqrt{\dfrac{b}{a}}\int_0^{\infty} e^{-\frac{1}{2}\left(ay^2 + \frac{b}{y^2}\right)}\dfrac{dy}{y^2}.$.
May I intrude and ask did you differentiate the change of variable x wrt to y? If you did, where did the minus sign go? If you didnt, then I dont understand whats happening :-)

5. Originally Posted by bugatti79
May I intrude and ask did you differentiate the change of variable x wrt to y? If you did, where did the minus sign go? If you didnt, then I dont understand whats happening :-)
In that change of variable, there is indeed a minus sign attached to the dy. But the limits of integration get switched: instead of going from 0 to ∞, the integral goes from ∞ to 0. When you switch the limits back again, that introduces a change of sign, which cancels with the other one.

6. Originally Posted by Opalg
In that change of variable, there is indeed a minus sign attached to the dy. But the limits of integration get switched: instead of going from 0 to ∞, the integral goes from ∞ to 0. When you switch the limits back again, that introduces a change of sign, which cancels with the other one.
Ok, how do the limits of integration get switched? Could an example be shown?

Thanks

7. Originally Posted by bugatti79
Ok, how do the limits of integration get switched? Could an example be shown?
When you make the substitution $x = \phi(y)$ in the definite integral $\int_a^b f(x)\, dx,$ it becomes $\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(y))\phi'(y)\,dy.$ In this case, $\textstyle x = \sqrt{\frac ba}\frac1y,$ and the limits of integration are that x goes from 0 to ∞. When x = 0, y = ∞; and when x = ∞, y = 0. So the dy-integral, instead of going from 0 to ∞, goes from ∞ to 0.

8. Originally Posted by Opalg
When you make the substitution $x = \phi(y)$ in the definite integral $\int_a^b f(x)\, dx,$ it becomes $\int_{\phi^{-1}(a)}^{\phi^{-1}(b)}f(\phi(y))\phi'(y)\,dy.$ In this case, $\textstyle x = \sqrt{\frac ba}\frac1y,$ and the limits of integration are that x goes from 0 to ∞. When x = 0, y = ∞; and when x = ∞, y = 0. So the dy-integral, instead of going from 0 to ∞, goes from ∞ to 0.
Ok, that makes sense. I have another question :-) At the start, how does he spot that x should be = (b/a)^1/2 * (1/y) in order to proceed?

Thanks

9. Under a change of variables, I literally wanted the two terms

$ax^2$ and $\dfrac{b}{x^2}$ to switch places. The change of variables

$x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$ does that.

10. Originally Posted by Danny
Under a change of variables, I literally wanted the two terms

$ax^2$ and $\dfrac{b}{x^2}$ to switch places. The change of variables

$x = \sqrt{\dfrac{b}{a}}\dfrac{1}{y}$ does that.
Ok, but why? I just dont see the connection. For example if I was given another similar example, I wouldnt know what to write x = to

11. Originally Posted by bugatti79
At the start, how does he spot that x should be = (b/a)^1/2 * (1/y) in order to proceed?
Sheer genius! You might deduce from the number of "Thanks" that he received for that solution that most of the other resident experts here had thought unsuccessfully about how to evaluate that integral. So you are not the only one to have found it difficult.