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## Homework Statement

Either by using the properties of convolution or directly from the definition, show that:

If

$$F(t)=\int^t_{-\infty} f (\tau) d \tau$$

then

$$(F * g) (t) = \int^t_{-\infty} (f * g) (\tau) d \tau$$

## Homework Equations

The convolution of ##f## with ##g## is given by:

$$(f*g)(t) = \int^{\infty}_{-\infty} f (\tau) g(t- \tau) d \tau$$

## The Attempt at a Solution

From definition:

$$(F * g) (t) = \int^{\infty}_{-\infty} \left( \left( \int^t_{-\infty} f (\tau) d \tau \right) . g(t-\tau) \right) d \tau$$

Switching the order of integration:

$$= \int^{t}_{-\infty} \left( \int^{\infty}_{-\infty} f (\tau) d \tau \right) d \tau . \int^{\infty}_{-\infty} g (t - \tau) d \tau$$

So, from here how can I show that this is equal to ##\int^t_{-\infty} (f*g) (\tau) d \tau##?

Because I must show that the expression above is equal to:

$$(F * g)(t) = \int^t_{-\infty} (f*g) (\tau) d \tau = \int^t_{-\infty} \left( \int^{\infty}_{-\infty} f (\tau) g(t-\tau) d \tau \right) d \tau$$

What properties of the convolution integral do I need to use?

Any help would be greatly appreciated.