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I am stuck on a question regarding finding the derivative $f'(0)$ of a function $f(x)$ defined as $f(x) = \int_{-\infty}^{e^{2x}} e^{\frac{-1}{2}(y-sin(x))^2}\, dy $ for $x \epsilon {\Bbb R}$.

I am vaguely aware that Leibniz' Integral Rule is the starting point for this question, but I don't know how to deal with the $-\infty$ in the lower bound of the integral. (https://en.wikipedia.org/wiki/Leibniz_integral_rule)

How do I compute this integral? Many thanks for any help given.

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  • $\begingroup$ You could split the integral ($\int_{-\infty}^0 + \int_0^{e^{2x}}$). Then you can just pull in the differentiation into the first integral and use Leibnitz on the second. $\endgroup$
    – amsmath
    Commented Sep 7, 2019 at 15:55
  • $\begingroup$ @amsmath I will try this out, thank you $\endgroup$
    – arnavlohe15
    Commented Sep 7, 2019 at 16:02

1 Answer 1

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We have that $$f(x) = \int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\mathrm d y.$$ Thus, $$f'(x) = \left(e^{2x}\right)'\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}[-\left(y-\sin x\right)](-\cos x)\mathrm d y=2e^{2x}\left[e^{-\frac{1}{2}\left(e^{2x}-\sin x\right)^2}\right]+\int_{-\infty}^{e^{2x}} e^{-\frac{1}{2}\left(y-\sin x\right)^2}\left(y-\sin x\right)(\cos x)\mathrm d y.$$

Now set $x=0$ to obtain $f'(0).$

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  • $\begingroup$ So I actually did derive something similar, but aren't there missing terms from this answer? $\endgroup$
    – arnavlohe15
    Commented Sep 7, 2019 at 16:01
  • $\begingroup$ @arnavlohe15 Oops, right. There were missing terms. $\endgroup$
    – Allawonder
    Commented Sep 7, 2019 at 16:08
  • $\begingroup$ how did you compute this integral? Can you show me the steps/explain the reasoning? $\endgroup$
    – arnavlohe15
    Commented Sep 7, 2019 at 16:09
  • $\begingroup$ @amsmath Perhaps you saw this before I edited. Is it still wrong? $\endgroup$
    – Allawonder
    Commented Sep 7, 2019 at 16:17
  • $\begingroup$ @arnavlohe15 I've made further editions. The first term comes from the chain rule applied to the integral. The second terms come from differentiating the integrand. $\endgroup$
    – Allawonder
    Commented Sep 7, 2019 at 16:18

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