Related to fourier transfrom figuring out when a function is odd or even

  • Thread starter SUDOnym
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  • #1
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I left a post up re. Fourier transforms and this query is on the same question:

having solved the heat equation in infinite domain in terms of U(alpha,t) we now want to inverse transform back to u(x,t):

[tex]U(\alpha,t)=\frac{2u_{0}}{\sqrt{2\pi}}\frac{sin\alpha}{\alpha}e^{-k^{2}\alpha^{2}t}[/tex]

[tex]\implies u(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}U(\alpha,t)e^{i\alpha x}d\alpha[/tex]

[tex]=\frac{u_{0}}{\pi}\int_{-\infty}^{\infty}\frac{\sin\alpha}{\alpha}e^{-i\alpha x}e^{-k^{2}\alpha^{2}t}d\alpha=\frac{u_{0}}{\pi}\int_{-\infty}^{\infty}\frac{\sin\alpha}{\alpha}\cos\alpha xe^{-k^{2}\alpha^{2}t}d\alpha-i\frac{u_{0}}{\pi}\int_{-\infty}^{\infty}\frac{\sin\alpha}{\alpha}\sin\alpha xe^{-k^{2}\alpha^{2}t}d\alpha[/tex]

My question here is:

in my notes he immediately says that in the final equality I have above that the sin(alpha)/alphacos(alpha)x term inside the integral is an even function whereas the sin(alpha)/alphasinalphax term is odd...how is it that he can so easily see which terms are odd and even and so evaluate the odd integral as zero?
 

Answers and Replies

  • #2
89
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I may not have been clear in my last post, question is how does he know that:

[tex]\frac{u_{0}}{\pi}\int_{-\infty}^{\infty}\frac{\sin\alpha}{\alpha}\cos\alpha xe^{-k^{2}\alpha^{2}t}d\alpha[/tex]

is even.

And how does he know that:

[tex]-i\frac{u_{0}}{\pi}\int_{-\infty}^{\infty}\frac{\sin\alpha}{\alpha}\sin\alpha xe^{-k^{2}\alpha^{2}t}d\alpha[/tex]

is odd?
 

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