- #1

- 89

- 1

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?