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March 9th, 2017, 09:54 AM   #1
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Complicated integral

Hello guys, I am working on an integral I need to solve, being of this type:
\begin{eqnarray}
I(z)=\int_{\lambda_{1}(z)}^{\lambda_{2}(z)}dx \arctan (bx -a)x^{\gamma}
\end{eqnarray}
where all $\gamma,b,a$ are real constants. The limits of integration are necessary, since after integrating I have to derive the expression respect to another variable that appears in the limits. Is there any approach to calculate this? I hope so because of the power law convolution with the arctan, maybe there is some analytical method for it. Thanks !
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March 9th, 2017, 10:27 AM   #2
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I am trying to use the arctan expansion

\begin{eqnarray}
\arctan(x)=\sum_{n=0}^{+\infty}\frac{x^{2n+1}}{(2n +1)!}
\end{eqnarray}
and calculate the value of the integral for a given n, then summing up only those relevant terms, am I allowed to do this?
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March 9th, 2017, 11:16 AM   #3
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The indefinite integral can be found analytically if $\gamma$ is a specific natural number. If you want the result, I suggest you use WolframAlpha.

If you need just I'(z), and not I(z), the problem is trivial, of course.

Last edited by skipjack; March 9th, 2017 at 11:25 AM.
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March 9th, 2017, 11:19 AM   #4
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Yes, as long as the sum is uniformly convergent. And, of course, a power series is uniformly convergent inside its radius of convergence. But I am not clear what you mean by "relevant terms".
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March 9th, 2017, 12:31 PM   #5
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I'm not sure I understand correctly. You are interested in computing $I'(z)$? If this is the case you don't need to integrate anything. Simply apply the fundamental theorem of calculus and the chain rule. Assuming that $F(\lambda) = \int_c^\lambda I'(\lambda(z)) d \z$ for some conveniently chosen lower limit and you have $I(z) = F(\lambda_1) - F(\lambda_2)$ and now taking a derivative you get
$$I'(z) = \arctan(b(\lambda_1)-a)\lambda_1^{\gamma}\cdot \frac{d\lambda_1}{dz} - \arctan(b(\lambda_2)-a)\lambda_2^{\gamma}\cdot \frac{d\lambda_2}{dz}$$
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Last edited by SDK; March 9th, 2017 at 12:34 PM.
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March 9th, 2017, 01:42 PM   #6
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Where was anything said about differentiating the integral?
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March 9th, 2017, 02:16 PM   #7
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The wording used was "after integrating I have to derive the expression respect to another variable that appears in the limits". It appears that "derive" means "differentiate" and "the expression respect" means "$I(z)$ with respect", etc.
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March 9th, 2017, 02:38 PM   #8
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Ok just to clarify things, the thing to compute is the derivative of I(z) respect to the variable z, which appears only in the limits of the integrals, so the integral in x needs to be performed before, the integral is in x, z is just a parameter appearing in the limits, for example, one of the limits is:

\begin{eqnarray}
\lambda_{1}=1/z + c
\end{eqnarray}
or similar.
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March 9th, 2017, 02:39 PM   #9
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Quote:
Originally Posted by Country Boy View Post
Yes, as long as the sum is uniformly convergent. And, of course, a power series is uniformly convergent inside its radius of convergence. But I am not clear what you mean by "relevant terms".
Those terms contributing more to the integral
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March 9th, 2017, 02:46 PM   #10
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As SDK explained, finding the derivative of $I(z)$ doesn't require that you first find $I(z)$.
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