I have no idea how to search this. Let's say I have two continuous functions on (0, c), f(t, x) and g(t, x), for which the following have no known closed forms:

[math]\int_{0}^{c}f(t,x)dt[/math]

[math]\int_{0}^{c}g(t,x)dt[/math]

If I write this:

[math]\int_{0}^{c}f(t,x)dt\int_{0}^{c}g(t,x)dt = \int_{0}^{c}h(t,x)dt[/math]

What can I say about h(t, x)? Can I define it in terms of f(t,x) and g(t,x)? Thanks

## Strange Calculus Question

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### Strange Calculus Question

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### Re: Strange Calculus Question

Hm. Smells a bit like it could be homework, so I'll mention that under sufficient smoothness assumptions (what you have there should be good enough, but someone check my sloppy thinking ) on f and g, you could apply the fundamental theorem of calculus and get an expression for h valid almost everywhere. But that still leaves you with an expression containing the integrals that have no closed form.

Are you looking to define h with that property from f and g, or just say what you can about h, given that it satisfies that property?

Are you looking to define h with that property from f and g, or just say what you can about h, given that it satisfies that property?

What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

### Re: Strange Calculus Question

This is question I've been curious about for ages. If h can be defined in terms of f and g, sweet.

http://aselliedraws.tumblr.com/ - surreal sketches and characters.

### Re: Strange Calculus Question

The convolution of f and g will give a function on (-c,c) with the desired property. If you want h to be defined on (0,c), you can rescale it fairly easily.

### Re: Strange Calculus Question

If you differentiate both sides, you get

[math]g(c,x)\int_{0}^{c}f(t,x)dt + f(c,x)\int_{0}^{c}g(t,x)dt = h(c,x)[/math]

[math]g(c,x)\int_{0}^{c}f(t,x)dt + f(c,x)\int_{0}^{c}g(t,x)dt = h(c,x)[/math]

she/they

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### Re: Strange Calculus Question

Sizik wrote:If you differentiate both sides, you get

[math]g(c,x)\int_{0}^{c}f(t,x)dt + f(c,x)\int_{0}^{c}g(t,x)dt = h(c,x)[/math]

Which you can just as easily re-write as:

[math]h(t, x) = g(t, x)\int_{0}^{t}f(t',x)dt' + f(t, x)\int_{0}^{t}g(t', x)dt'[/math]

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### Re: Strange Calculus Question

Nitrodon wrote:The convolution of f and g will give a function on (-c,c) with the desired property. If you want h to be defined on (0,c), you can rescale it fairly easily.

Hm. So integration turns convolutions into products. That nugget will probably come in handy, one day.

What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

### Re: Strange Calculus Question

Interesting, thanks for the responses.

http://aselliedraws.tumblr.com/ - surreal sketches and characters.

### Re: Strange Calculus Question

z4lis wrote:Nitrodon wrote:The convolution of f and g will give a function on (-c,c) with the desired property. If you want h to be defined on (0,c), you can rescale it fairly easily.

Hm. So integration turns convolutions into products. That nugget will probably come in handy, one day.

It's super-handy in signal processing.

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