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How would I find the derivative of

$$\int_{-\infty}^{-x} f(-u) du$$

with respect to $x$? Am I supposed to use the Fundamental Theorem of Calculus?

I try to let another variable $m=-u$ but I end up getting

$$\int_{x}^{\infty}f(m)dm $$

and I have no idea what to do with that.

Thanks! :)

Yuxiao Xie
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Ronny D
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    Yes you are. The Chain Rule will also be used. – André Nicolas May 21 '16 at 05:10
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    I try to let another variable m = -u but then my limits become really weird and I get stuck. I keep getting the second integral in the newest edit of my post – Ronny D May 21 '16 at 05:15
  • Going off your last part with $f(m)$, try swapping bounds and then applying Fundamental Theorem. – Leonidas Lanier May 21 '16 at 05:21
  • If I negative the integral to get infinity as the lower bound, does the Fundamental Theorem work with that? That's what I'm so confused about >.<" – Ronny D May 21 '16 at 05:24
  • I'm inclined to say so. I've seen tons of different lower bounds where Fundamental Theorem was applied (i.e. $\forall x\in\Bbb{R}$) and I don't see that being any different for infinity. If I'm wrong someone correct me. – Leonidas Lanier May 21 '16 at 05:27
  • The answer will be $(-1)f(-(-x))$. The $-1$ in front comes from differentiating $-x$. More simply $-f(x)$. You can get the same thing out of your suggested substitution. – André Nicolas May 21 '16 at 05:29
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    Related: http://math.stackexchange.com/questions/83971/is-there-a-fundamental-theorem-of-calculus-for-improper-integrals – Leonidas Lanier May 21 '16 at 05:31
  • Alright I think I get it now! Thank you both so much!!! – Ronny D May 21 '16 at 05:43

2 Answers2

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The given integral is, explicitly, $$ g(x)=\lim_{t\to\infty}\int_{-t}^{-x} f(-u)\,du $$ which is assumed to exist for every $x$. By doing $u=-v$, we get $$ \int_{-t}^{-x} f(-u)\,du=-\int_t^x f(v)\,dv= -\int_{t}^{x_0}f(v)\,dv-\int_{x_0}^x f(v)\,dv $$ where $x_0$ is some fixed value.

Then we have $$ g(x)=-\int_{x_0}^xf(v)\,dv-\lim_{t\to\infty}\int_{t}^{x_0}f(v)\,dv $$ Since the limit is a constant not depending on $x$, the derivative is $$ g'(x)=-f(x) $$ by the fundamental theorem.

egreg
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Let $\phi(u) = f(-u)$, and $h(t) = \int_{-\infty}^t \phi(y) dy$. Then $h'(t) = \phi(t) = f(-t)$.

Let $m(x) = -x$, then $m'(x) = -1$.

Now let $g(x) = \int_{-\infty}^{-x} \phi(y) dy = h(-x) = (h \circ m) (x)$.

Then $g'(x) = h'(m(x)) m'(x) = - f(x)$.

copper.hat
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