How to calculate the following integral: $$\int_{-\infty}^{\infty} \exp{\left({-a x^2 -\frac{b}{x^2}}\right)} \,dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}.$$
Asked
Active
Viewed 111 times
2
-
1You can complete in the square. – Zaid Alyafeai Apr 15 '17 at 01:28
-
$$\int_{-\infty}^{+\infty}f\left(x-\frac{a}x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x): \mathrm{d}x$$ – Zaid Alyafeai Apr 15 '17 at 01:39
-
@user1952009, see my answer. – Zaid Alyafeai Apr 15 '17 at 01:49
-
If $a<0$, the integral diverges anyway. – Chappers Apr 15 '17 at 01:49
-
@ZaidAlyafeai Greatly thanks – 346699 Apr 15 '17 at 01:58
1 Answers
2
$$\left({-a x^2 -\frac{b}{x^2}}\right) = -\left({a x^2 -2\sqrt{ab}+\frac{b}{x^2}}\right)-2\sqrt{ab} =-a\left({x -\frac{\sqrt{b/a}}{x}}\right)^2-2\sqrt{ab} $$
Then using that
$$\int_{-\infty}^{+\infty}f\left(x-\frac{c}x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x$$
With $f(x) = e^{-ax^2}$ and $c=\sqrt{b/a}$
$$I =\int_{-\infty}^{\infty} \exp{\left({-a x^2 -\frac{b}{x^2}}\right)} \,dx =e^{-2\sqrt{ab}} \int_{-\infty}^{\infty}e^{-ax^2}\,dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}$$
Reference
Zaid Alyafeai
- 14,343