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The Taylor series of the function

$$f(x) = \int_{1}^{\sqrt{x}} \ln(xy)+ y^{3x} dy + e^{2x}$$

at the point $x = 1$ is

$$e^2 + (x-1)\left(2e^2+\frac{1}{2}\right) + \frac{(x-1)^2}{2}\left(4e^2+\frac{7}{4}\right)$$

which I calculated using the Leibniz rule.

  • How can I estimate the remainder term of second order for f(2) ? (The second derivate is already very complicated).

  • Is there a method to calculate higher derivatives of parameter integrals easier than simply applying the Leibniz rule repeatedly?

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Try to use the integral form for the remainder

$$ R_{n+1}(x)=\int_{a}^{x}\frac{f^{(n+1)}(t)}{n!}(x-t)^n\,dt\,. $$

Here is a related technique. Note that, in your case $a=1$.

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  • $\begingroup$ I know this representation. But the second derivate is very complicated. Is there a method to estimate the derivate in such a case ? $\endgroup$ – Peter Feb 5 '14 at 15:49
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I wonder if there is not a typo somewhere. The integral in your post can be computed but the Taylor expansion of the overall function, built at $x=1$, is (hoping that I did not make any mistake) $$e^2+2 e^2 (x-1)+\left(\frac{5}{8}+2 e^2\right) (x-1)^2+\left(\frac{1}{24}+\frac{4 e^2}{3}\right) (x-1)^3+O\left((x-1)^4\right)$$

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  • $\begingroup$ Yes, there is indeed a typo. It should be $ln(xy)+y^{3x}$ instead of $ln(xy)y^{3x}$. I fixed it, sorry. $\endgroup$ – Peter Feb 5 '14 at 15:48

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