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Hello,

 

Given that \(f(x)\) is a function that satisfies \(\displaystyle\int_{-\infty}^{\infty} e^{tx}f(x)\,dx = \sin^{-1}\left(t - \sqrt{\frac12}\right)\), for all possible values of \(t\), compute the doubly unbounded integral:

\(\int_{-\infty}^{\infty} xf(x)\,dx. \)

 

Thank you so much and enjoy!

 

JS

 Dec 26, 2021

Best Answer 

 #3
avatar+32828 
+2

I get \(\sqrt2\) as follows:

 Dec 28, 2021
 #1
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+1

Since the domain of arcsin is [-1,1], you know that \(1/\sqrt{2} \le t \le 1 + 1/\sqrt{2}\)

 

Also, by definition of improper integrals,

\(\int_{-\infty}^\infty e^{tx} f(x) dx = \lim_{a \to -\infty} \int_a^0 e^{tx} f(x) dx + \lim_{b \to \infty} \int_0^b e^{tx} f(x) dx = \sin^{-1} \left( t - \frac{1}{\sqrt{2}} \right)\)

\(\int_{-\infty}^\infty xf(x) dx = \lim_{c \to -\infty} \int_c^0 f(x) dx + \lim_{d \to \infty} \int_0^d xf(x) dx\)

 

With some more bounding, you can get that \(\int_{-\infty}^\infty xf(x) dx = \dfrac{\pi \sqrt{2}}{4}\)

 Dec 27, 2021
 #2
avatar+45 
0

Hmm... I'm getting \(\frac{\pi\sqrt{2}}{2}\)...

 

Is anyone getting another answer?

jsaddern  Dec 27, 2021
 #5
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Hi Guest,

I do believe that Alan's answer is the correct one but I would like to understand what you have tried to do.

 

I think you have accidentally omitted an x in the second row but that is no bigf deal

 

When you say 'and with more bounding'  I have no idea what you mean.

Melody  Dec 28, 2021
 #3
avatar+32828 
+2
Best Answer

I get \(\sqrt2\) as follows:

Alan Dec 28, 2021
 #4
avatar+115897 
0

That is a great solution Alan, which means I can easily understand it.  Thanks 

Melody  Dec 28, 2021

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