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I would like to differentiate a function of the type $\int_x^\infty f(x, t) dt$ with respect to $x$ ($f$ real or complex valued). Does differentiation under the integral sign apply? What are better methods to differentiate this?

Thanks!

Steven
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1 Answers1

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One may use Leibniz integral rule

$$ \frac{\mathrm{d}}{\mathrm{d}x} \left (\int_{a(x)}^{b(x)} f(x,t)\,\mathrm{d}t \right )=b'(x) \cdot f\big(b(x),x\big) \,-\,a'(x)\cdot f\big(a(x),x \big)+ \int_{a(x)}^{b(x)}\frac{ \partial f}{ \partial x}(x,t)\,\mathrm{d}t $$

giving here, with appropriate hypotheses,

$$ \frac{\mathrm{d}}{\mathrm{d}x} \left (\int_x^{+\infty} f(x,t)\,\mathrm{d}t \right )=-\, f\big(x,x \big)+ \int_x^{+\infty}\frac{ \partial f}{ \partial x}(x,t)\,\mathrm{d}t . $$

Olivier Oloa
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  • How did $b'(x)\cdot f(b(x), x)$ vanish here, considering that the upper bound $b(x)=\infty$? – Milos Dec 05 '19 at 16:58
  • Please have a look at the hyperlink above. – Olivier Oloa Dec 07 '19 at 11:33
  • I did, but didn't get how the part with $b'(x)$ vanish. – Milos Dec 07 '19 at 13:16
  • Here $b(x):=M$ which is a constant with respect to $x$ (thus $b'(x)=0$), but this is true for any constant $M$ as great as we want ($M>M_0$) giving the second identity above. – Olivier Oloa Dec 07 '19 at 14:29
  • Can we treat infinity as a constant with respect to a variable, as $\infty$ is not a value? – Milos Dec 07 '19 at 14:46
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    Yes if we can switch limits : an improper integral is defined with a limit, differentiating is defined with a limit, so if one is allowed to switch the order of limits, then we are done... that's why I had written 'with appropriate hypotheses'... – Olivier Oloa Dec 08 '19 at 21:22
  • Do you happen to know a reference for the proof of the dominated convergence theorem-corollary result? – FShrike Apr 16 '22 at 16:46