+0

# Computing integral unbounded on both sides!

0
76
5
+45

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

#3
+32828
+2

I get $$\sqrt2$$ as follows:

Dec 28, 2021

#1
+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
+45
0

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

#5
+115897
0

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
+32828
+2
I get $$\sqrt2$$ as follows: