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I have $F(x)$, a non-negative function, and a definite integral: $$\int_{0}^{\infty}F(-x)dx$$

How do I use substitution of variable?

If I do this with $u = -x$, then $dx = -du$ and with new limits of integration: $0$ and $-\infty$ resulting in: $$-\int_{-\infty}^{0}F(u)du$$ Now, I have a wrong answer because $F$ is non-negative, but the integral will be negative.

What's missing here?

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Your new integration limits over $u$ are reversed: $$\int_0^\infty F(-x)\,\text{d}x =-\int_0^{-\infty} F(u)\,\text{d}u =\int_{-\infty}^0 F(u)\,\text{d}u.$$

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  • $\begingroup$ So I have to put the new limits where they were and then reverse them because of the negative sign? I want to understand the passage from the second step to the third. $\endgroup$ – falecomdino Apr 30 '18 at 12:23
  • $\begingroup$ Yes, you put the new limits where they were. The next step is a property of definite integrals: reversing the integration limits actually flips the sign of the whole integral. $$\int_a^b =-\int_b^a.$$ $\endgroup$ – Rócherz Apr 30 '18 at 12:25

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