Is $(-1+i)\log(2e^{it}+i)$ same as $\frac{1}{2}\left((2+2i)\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})\right)$?

Question

Is $\displaystyle(-1+i)\log(2e^{it}+i)$ the same as $\displaystyle\frac{1}{2}\left((2+2i)\;\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})\right)$?
WolframAlpha shows that they are same, but this page on W|A shows FALSE so they are not same.

Advice: use www.tinyurl.com , or any other site of the kind, to shorten such large addresses, otherwise it gets messed up.

joriki · Accepted Answer · 2012-08-25 16:43:27Z

up vote 3 down vote accepted

Why did you obfuscate an assumption about an integral behind a WolframAlpha link and claimed that the linked page shows something else?

These two antiderivative need not be equal because they can differ by an additive constant.

You can let WolframAlpha subtract them to find the constant; the result shows that there are also issues with the multivaluedness of the functions involved.

answered Aug 25 '12 at 16:43

joriki
92.5k477167

	Why did I what? What assumption? – alvoutila Aug 27 '12 at 12:24
	@alvoutila: You included a link to WolframAlpha, claiming that it "shows they are the same". It shows nothing of the kind; it shows that WolframAlpha returns a certain antiderivative for a certain indefinite integral, but you mentioned none of that or any of the steps that you took in incorrectly concluding that this is equivalent to showing that these two expressions are the same. That's very misleading, and unnecessarily so; you could have described directly what the WolframAlpha link shows, and what you concluded from that and why. – joriki Aug 27 '12 at 12:49
	well this link [WolframAlpha][1] shows that $(-1+i)\log(2e^{it}+i)$ and $\frac{1}{2}(2+2i)\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})$"... is equivalent for restricted values of $t$..." [1]: wolframalpha.com/input/?i=integrate%28%20%28-2e%5E%28it%29-2ie%5E‌%28it%29%29/%282e%5E%28i*t%29%2bi%29,%20t%29 – alvoutila Aug 27 '12 at 16:18

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Is $(-1+i)\log(2e^{it}+i)$ same as $\frac{1}{2}\left((2+2i)\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})\right)$?

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Is $(-1+i)\log(2e^{it}+i)$ same as $\frac{1}{2}\left((2+2i)\tan^{-1}(2e^{it})-(1-i)\log(1+4e^{2it})\right)$?

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