+72 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
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 \(\int_{-\infty}^\infty xf(x) dx = \dfrac{\pi \sqrt{2}}{4}\)
+72 Hmm... I'm getting \(\frac{\pi\sqrt{2}}{2}\)...
Is anyone getting another answer?
+118687 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.
