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Post by mondo on Jun 18, 2023 6:34:34 GMT
On page 437 there the first equation represents a definite integral $\int_{-\infty}^{\infty}(\frac{dx}{(x^2+1)^2})$ I wonder if this can be solved somehow without knowing trigonometric identities? I tried this way:
$z = x^2 + 1$; $dz = 2xdx$; $dx = \frac{dz}{2x}$ hence $\frac{1}{2x}\int_{-\infty}^{\infty} \frac{1}{z^2}dz = -(\frac{1}{2xz} + C) = -\frac{1}{2x(x^2+1)} + C$. But now I am stuck as this tends to 0 from both sides so I am unable to calculate limits.
Vasco do you have some idea?
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Post by Admin on Jun 18, 2023 12:17:11 GMT
Mondo
As it says on page 436 at the beginning of the last paragraph we can do this using partial fractions. If you can't work it out, see my answer to Exercise 14 of chapter 8.
Your attempt above should be written as follows
$\displaystyle\int_{-\infty}^{\infty}\frac{dz}{2xz^2}=\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, but this doesn't help.
Vasco
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Post by mondo on Jun 18, 2023 18:45:58 GMT
Mondo As it says on page 436 at the beginning of the last paragraph we can do this using partial fractions. If you can't work it out, see my answer to Exercise 14 of chapter 8. Your attempt above should be written as follows $\displaystyle\int_{-\infty}^{\infty}\frac{dz}{2xz^2}=\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, but this doesn't help. Vasco Yes your solution to exercise 14 substitutes $x = \tan(\theta)$ and then applying a sequence of trigonometric identities it leads to the correct result. However how did you know, to substitute just $\tan(\theta)$ for $x$ and not say $cos(x)$? Also, $\displaystyle\int_{-\infty}^{\infty}\frac{dz}{2xz^2}=\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, but this doesn't help. How did you get $\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, it doesn't look correct. The integral of $\int_{-\infty}^{\infty}\frac{dz}{2xz^2}$ is just $\frac{-1}{2xz}$ |
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Post by Admin on Jun 18, 2023 18:55:46 GMT
Mondo
You are integrating as if $x$ is constant but it isn't. $x=\sqrt{z-1}$
Vasco
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Post by Admin on Jun 18, 2023 21:41:34 GMT
Mondo As it says on page 436 at the beginning of the last paragraph we can do this using partial fractions. If you can't work it out, see my answer to Exercise 14 of chapter 8. Your attempt above should be written as follows $\displaystyle\int_{-\infty}^{\infty}\frac{dz}{2xz^2}=\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, but this doesn't help. Vasco Yes your solution to exercise 14 substitutes $x = \tan(\theta)$ and then applying a sequence of trigonometric identities it leads to the correct result. However how did you know, to substitute just $\tan(\theta)$ for $x$ and not say $cos(x)$?
Also, $\displaystyle\int_{-\infty}^{\infty}\frac{dz}{2xz^2}=\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, but this doesn't help. How did you get $\int_{-\infty}^{\infty}\frac{dz}{2\sqrt{z-1}z^2}$, it doesn't look correct. The integral of $\int_{-\infty}^{\infty}\frac{dz}{2xz^2}$ is just $\frac{-1}{2xz}$[/quote] |
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