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 March 9th, 2017, 08:54 AM #1 Member   Joined: Dec 2016 From: - Posts: 54 Thanks: 10 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 !
 March 9th, 2017, 09:27 AM #2 Member   Joined: Dec 2016 From: - Posts: 54 Thanks: 10 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?
 March 9th, 2017, 10:16 AM #3 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1384 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 10:25 AM.
 March 9th, 2017, 10:19 AM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 680 Yes, $\int (\sum f_n(x))g(x)dx= \sum\int f_n(x)g(x)dx$ 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". Thanks from nietzsche
 March 9th, 2017, 11:31 AM #5 Senior Member   Joined: Sep 2016 From: USA Posts: 114 Thanks: 45 Math Focus: Dynamical systems, analytic function theory, numerics 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 $$I'(z) = \arctan(b(\lambda_1)-a)\lambda_1^{\gamma}\cdot \frac{d\lambda_1}{dz} - \arctan(b(\lambda_2)-a)\lambda_2^{\gamma}\cdot \frac{d\lambda_2}{dz}$$ Thanks from nietzsche Last edited by SDK; March 9th, 2017 at 11:34 AM.
 March 9th, 2017, 12:42 PM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,649 Thanks: 680 Where was anything said about differentiating the integral?
 March 9th, 2017, 01:16 PM #7 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1384 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.
 March 9th, 2017, 01:38 PM #8 Member   Joined: Dec 2016 From: - Posts: 54 Thanks: 10 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.
March 9th, 2017, 01:39 PM   #9
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Quote:
 Originally Posted by Country Boy Yes, $\int (\sum f_n(x))g(x)dx= \sum\int f_n(x)g(x)dx$ 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

 March 9th, 2017, 01:46 PM #10 Global Moderator   Joined: Dec 2006 Posts: 17,919 Thanks: 1384 As SDK explained, finding the derivative of $I(z)$ doesn't require that you first find $I(z)$. Thanks from nietzsche

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