Gear300
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How would I integrate e^(-pi^2*x^2) from -infinite to infinite? I know that the integral of e^(-x^2) from -infinite to infinite is sqr(pi), in which sqr is square root.
How to Integrate e^(-pi^2*x^2) from -infinite to infinite?
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Science Advisor
Homework Helper
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In general once can prove that
[tex]\int_{-\infty}^{\infty} e^{-\alpha x^2} \, \mathrm{d}x = \sqrt{\frac{\pi}{\alpha}},[/tex]
for [itex]\alpha > 0[/itex] (and even [itex]\alpha \in \mathbb{C}, \operatorname{Re}(\alpha) > 0[/itex]).
Then if you take [itex]\alpha = 1[/itex] or [itex]\alpha = \pi^2[/itex] you will get either of the integrals in your post.
[tex]\int_{-\infty}^{\infty} e^{-\alpha x^2} \, \mathrm{d}x = \sqrt{\frac{\pi}{\alpha}},[/tex]
for [itex]\alpha > 0[/itex] (and even [itex]\alpha \in \mathbb{C}, \operatorname{Re}(\alpha) > 0[/itex]).
Then if you take [itex]\alpha = 1[/itex] or [itex]\alpha = \pi^2[/itex] you will get either of the integrals in your post.
Gear300
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I see...thank you.
Schrodinger's Dog
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Actually, I can see in this case that a=b so b-a=0 and therefore
[tex]\int_a^b e^{-\pi x^2}\,dx \rightarrow \int_{-\infty}^{\infty} e^{-\pi x^2}\,dx=1[/tex]
But that's probably beside the point.
[tex]\int_a^b e^{-\pi x^2}\,dx \rightarrow \int_{-\infty}^{\infty} e^{-\pi x^2}\,dx=1[/tex]
But that's probably beside the point.