Fourier transform and convolution, dirac function

nightworrier
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Hi everyone,
I uploaded a solution about Fourier transform. At the solution of this problem, it states that make convolution. But i tried to do convolution but my result is not same with this result. When you do the convolution with 2.10 and 2.11, is the result 2.13 correct ? How is it done ? :S I am so confused.
 

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i really don't understand the basis of this problem

but from what you have posted if E2.9 to E2.12 is the way it is then E2.13 should be right

did you do the convolution of 2.10 with 2.11?
 
I try to do the convolution. But i think at E2.13 there shouldn't be 1/4. Maybe i do something wrong.
I said E2.10 x(f)
Then E2.11 y(f)

x(f)*y(f)=int(x(z).y(f-z)dz) right ? this is the convolution
so i put E2.11 into the integral. And i subsitute f to f-z the formula becomes
\frac{1}{2}int(\frac{1}{\frac{1}{2}+j2.pi.f}.[dirac(f-50-z)+dirac(f+50-z)]dz

then f-50-z=0 df=dz
z=f-50
i computed the integral than put z into these formula. But 1/4 not coming from there. Maybe i did something wrong
 
oops i don't get that 1/4 either, may be it is some typo in that sheet
 
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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