Orthogonal Functions | Homework Statement

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Homework Statement




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The Attempt at a Solution



All functions orthogonal to 1 result in the fact that: \int_a^b f(t)\ \mbox{d}t =0

Now the extra condition is that f must be continous. (because of the intersection).

But where does the fact that f(a)=f(b)=0 comes from? And why look at the deratives?
 
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Remember way back in first year calc, when you learned that to do that integral you find an antiderivative F(x) and evaluate F(b)-F(a). This is that same problem in disguise.
 
Well I thought of this: \int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0
 
dirk_mec1 said:
Well I thought of this: \int_a^b \int_a^t f(s)\ \mbox{d}s \mbox{d}t =0

Fine. What are you going to do with it? Why don't you just define <br /> F(x)=\int_a^x f(s)\ \mbox{d}s<br />
What are some of the properties of F(x)?
 
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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