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August 21

How to integrate it

How to do this indefinite integration?

\int x\tan xdx

Sayan19ghosh99 (talk) 13:46, 21 August 2017 (UTC)[reply]

The antiderivative is not expressible as an elementary function. But you can always write out the integrand's Taylor series and then integrate term-by-term to get a series representation for it. --Deacon Vorbis (talk) 14:21, 21 August 2017 (UTC)[reply]

https://www.wolframalpha.com/input/?i=integrate+x+tan+x+dx Bo Jacoby (talk) 09:11, 24 August 2017 (UTC).[reply]

... and we have a brief article on Polylogarithmic function. Should it be expanded to include this type of application? Dbfirs 16:16, 24 August 2017 (UTC)[reply]

You're mixing it up with the polylogarithm. --Deacon Vorbis (talk) 16:38, 24 August 2017 (UTC)[reply]

... so I am! It was Wolfram's use of "Polylogarithm function" that misled me. I should have read both articles more carefully! Dbfirs 07:03, 27 August 2017 (UTC)[reply]

Solving linear differential equations with constant coefficients

Can I solve all linear differential equations with constant coefficient and nonzero right hand side by factorizing method?

Sayan19ghosh99 (talk) 13:47, 21 August 2017 (UTC)[reply]

In theory you can solve any homogeneous (RHS = 0) LDE with constant coefficients by factorizing its characteristic polynomial. In practice finding the exact values of the roots of a high degree polynomial may be difficult or impossible. To find the general solution to an inhomogeneous (RHS != 0) LDE with constant coefficients you have to find a particular integral as well as solving the homogeneous version of the LDE. Gandalf61 (talk) 14:12, 21 August 2017 (UTC)[reply]

I assume by "factorizing method" you mean that you factorize the differential operator. As Gandalf61 notes above, this is difficult to do in practice for high order equations. Once you have overcome this difficulty, you can then solve for any continuous RHS using repeated application of the integrating factor method (but there may not be closed form expressions for some of the integrals you need to calculate). —Kusma (t·c) 14:56, 23 August 2017 (UTC)[reply]

The homogenous equation is $\left(\sum _{k=0}^{n}a_{k}\left({\frac {d}{dx}}\right)^{k}\right)y=0$ where $a_{n}=1$ . If $y=e^{\lambda x}$ is a solution, then $\lambda$ satisfies the algebraic equation $\sum _{k=0}^{n}a_{k}\lambda ^{k}=0$ . This equation is solved numerically by, say, the Durand-Kerner method, giving the roots $\lambda _{1},\cdots ,\lambda _{n}$ such that $\sum _{k=0}^{n}a_{k}\lambda ^{k}=\prod _{k=1}^{n}(\lambda -\lambda _{k})$ . When the roots are all different from one another the general solution is $y=\sum _{k=1}^{n}b_{k}e^{\lambda _{k}x}$ where $b_{1},\cdots ,b_{n}$ are arbitrary constants. Bo Jacoby (talk) 17:30, 24 August 2017 (UTC).[reply]