Double integral fail, works on TI
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04-01-2020, 09:58 AM
Post: #5
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RE: Double integral fail, works on TI
Could it be because
∫(e^((-x)*y)/(x^2-y^2),y) returns (Ei(-x^2-x*y)*(e^x^2)^2-Ei(x^2-x*y))/(2*x*e^x^2) replacing y with x or -x results in Ei(0) which equals -infinity? |
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Messages In This Thread |
Double integral fail, works on TI - lrdheat - 04-01-2020, 02:23 AM
RE: Double integral fail, works on TI - lrdheat - 04-01-2020, 02:25 AM
RE: Double integral fail, works on TI - johnet123 - 04-01-2020, 08:06 AM
RE: Double integral fail, works on TI - CyberAngel - 04-01-2020, 09:20 AM
RE: Double integral fail, works on TI - roadrunner - 04-01-2020 09:58 AM
RE: Double integral fail, works on TI - froehlic - 04-01-2020, 01:44 PM
RE: Double integral fail, works on TI - lrdheat - 04-01-2020, 05:29 PM
RE: Double integral fail, works on TI - Albert Chan - 04-01-2020, 09:20 PM
RE: Double integral fail, works on TI - parisse - 04-02-2020, 12:33 PM
RE: Double integral fail, works on TI - CyberAngel - 04-02-2020, 12:46 PM
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