Simple integration question involving-infty subscript

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simple integration question involving-infty "subscript"

Homework Statement



Been reading about signals, but my calculus skills have rusted (or never has been all that good in the first place).

So ...

Homework Equations



Why does x(t) = \int^t_{-\infty} x'(\tau) \,d\tau ?

The Attempt at a Solution



You will end up with x(\tau)|^t_{-\infty} = x(t) - x({-\infty}). Right?

So x({-\infty}) = 0 for all functions?

Been looking on the web, but I have no idea how to google this.
 
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Salt said:

Homework Statement



Been reading about signals, but my calculus skills have rusted (or never has been all that good in the first place).

So ...

Homework Equations



Why does x(t) = \int^t_{-\infty} x'(\tau) \,d\tau ?

The Attempt at a Solution



You will end up with x(\tau)|^t_{-\infty} = x(t) - x({-\infty}). Right?

So x({-\infty}) = 0 for all functions?

Been looking on the web, but I have no idea how to google this.
No, you can't do this. Infinity is not a number that you can substitute into a function. Your integral is one type of improper integral. To evaluate an integral like this, you need to work with a limit, like so:
x(t) = \int^t_{-\infty} x'(\tau) \,d\tau = \lim_{a \to -\infty} \int_a^t x'(\tau) \,d\tau
 


Hmm ...

Looks like my calculus really does suck.

It was just written "like that" in the book. It probably assumes that I know how to solve it. ><

Thanks.
 
Last edited:
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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