Skip to main content
MathOverflow

Return to Question

Couldn't downvote without editing it first.
Source Link
Nemo
Nemo
  • 5.6k
  • 2
  • 29
  • 40

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.

Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$$$\int_0^{\infty}x^2e^{-x^2}\frac{{dx}}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.

Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.

Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{{dx}}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$

edited tags
Link
T. Amdeberhan
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217
Source Link
T. Amdeberhan
T. Amdeberhan
  • 43.2k
  • 5
  • 57
  • 217

Is there a transformation or a proof for these integrals?

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.

Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$