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Given $$ \int^{\infty}_{-\infty}e^{tx}f(x)dx = \sin^{-1}\bigg(t-\sqrt{\frac{1}{2}}\bigg),$$

then what is the value of $$ \bigg(\int^{\infty}_{-\infty}xf(x)dx\bigg)^2\quad ?$$

What I tried:

$$I(t)=\int^{\infty}_{-\infty}e^{tx}f(x)dx=\sin^{-1}\bigg(t-\sqrt{\frac{1}{2}}\bigg)$$

$$I'(t)=\int^{\infty}_{-\infty}e^{tx}tf(x)dx=\frac{1}{\sqrt{1-\bigg(t-\sqrt{\frac{1}{2}}\bigg)^2}}$$

I do not know how to solve it.

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jacky
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1 Answers1

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If you take derivative w.r.t. to $t$, then $$I'(t)=\int^{\infty}_{-\infty}e^{tx}xf(x)dx=\frac{1}{\sqrt{1-\bigg(t-\sqrt{\frac{1}{2}}\bigg)^2}}.$$ Now simply let $t=0$.

Mini
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    This (Liebnitz integral rule,https://en.wikipedia.org/wiki/Leibniz_integral_rule ) is also called "Feynman's trick". – GEdgar Mar 03 '20 at 14:49