We know that : $$\int_0^∞ e^{-x^2} dx = \frac {\sqrt{π}}{2}$$ $$\int_0^∞ e^{-x^2-\frac {a^2}{x^2}} dx = \frac {\sqrt{π}}{2}e^{-2a}$$ Both the above results can be easily proved by integration under integration. I was wondering if it can be extended to the general integral $\int_0^∞ e^{-f(x^2)} dx$. The result is surely $\frac {\sqrt{π}}{2}F$ where $F$ is a constant function. But I am unable to relate $F$ with $f$. Any suggestions are welcome.

  • $\begingroup$ How is the first result proved by differentiation under the integral? $\endgroup$ – Yuriy S Nov 11 '18 at 10:08
  • $\begingroup$ @YuriyS math.stackexchange.com/questions/390850/… $\endgroup$ – Botond Nov 11 '18 at 10:29
  • $\begingroup$ That citation mentions just $x^2.f(b)$ not $f(x^2)$. $\endgroup$ – Awe Kumar Jha Nov 11 '18 at 10:32
  • $\begingroup$ @Botond, in this question the constant $\sqrt{\pi}/2$ is not found, which is the point the author makes. Ah, never mind, I see the answer. Thanks $\endgroup$ – Yuriy S Nov 11 '18 at 10:36

Here's a little list with examples of the integrals in the form:

$$\int_0^\infty e^{-f(x^2)} dx$$

Gamma function related integrals:

$$\int_0^\infty e^{-x^2} dx =\Gamma \left( \frac{3}{2}\right)=\frac{1}{2} \Gamma \left( \frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}$$

$$\int_0^\infty e^{-x^4} dx =\Gamma \left( \frac{5}{4}\right)$$

$$\int_0^\infty e^{-x^{2n}} dx =\Gamma \left( \frac{2n+1}{2n}\right)$$

Bessel function related integrals:

$$\int_0^\infty e^{-a \cosh x} dx=K_0 (a)$$

$$\int_0^\infty e^{-a (1+\frac{4}{3} x^2) \sqrt{1+\frac{1}{3} x^2}} dx=\frac{1}{\sqrt{3}}K_{1/3} (a)$$

Many more examples are possible.

Trivial cases:

$$\int_0^\infty e^{-\ln (1+x^2)} dx=\int_0^\infty \frac{1}{1+x^2} dx=\frac{\pi}{2}$$

$$\int_0^\infty e^{-\frac{3}{2}\ln (1+x^2)} dx=\int_0^\infty \frac{1}{(1+x^2)^{3/2}} dx=1$$

$$\int_0^\infty e^{-\ln (1+x^4)} dx=\int_0^\infty \frac{1}{1+x^4} dx=\frac{\pi}{2\sqrt{2}}$$

$$\int_0^\infty e^{-\ln (\cosh x)} dx=\int_0^\infty \frac{1}{\cosh x} dx=\frac{\pi}{2}$$

And so on.

What this list is intended to show is that there's no general method for finding the closed forms for such integrals. They need to be dealt with on case by case basis. Sometimes generalization is possible, sometimes not.

An example of such a general theorem can be seen in this answer, which allows us to make an infinite number of integrals giving the same value.

  • $\begingroup$ Yet all the above mentioned results can be expanded so that the factor $\frac {√π}{2}$ may appear. $\endgroup$ – Awe Kumar Jha Nov 11 '18 at 13:52
  • $\begingroup$ @AweKumarJha, sure they can, but why would we need to do that? It's just a number $\endgroup$ – Yuriy S Nov 11 '18 at 13:59

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