Problem with integral

[rprosperi]

(12-16-2019 04:41 AM)Stevetuc Wrote:  

(12-15-2019 10:19 PM)rprosperi Wrote:  If you have access to Wolfram, and you trust it more than the Prime, why use the Prime at all? Just curious...

The purpose of reporting issues with a product to r&d is to improve the product. Checking a result against other sources is normal engineering practice.
Your question seems to imply you would be happy with this result:
((e^(1/2*τ))^2*(e^(i/2*τ))^2*tan(τ/4)^2+(e^(1/2*τ))^2*(e^(i/2*τ))^2+(-1-i)*(e^(1/2*τ))^2*e^(i/2*τ)*tan(τ/4)^2+(2+2*i)*(e^(1/2*τ))^2*e^(i/2*τ)*tan(τ/4)+(1+i)*(e^(1/2*τ))^2*e^(i/2*τ)+(−i)*(e^(1/2*τ))^2*tan(τ/4)^2+(−i)*(e^(1/2*τ))^2+i*(e^(i/2*τ))^2*tan(τ/4)^2+i*(e^(i/2*τ))^2+(1+i)*e^(i/2*τ)*tan(τ/4)^2+(2+2*i)*e^(i/2*τ)*tan(τ/4)+(-1-i)*e^(i/2*τ)-tan(τ/4)^2-1)/((2+2*i)*e^(1/2*τ)*e^(i/2*τ)*tan(τ/4)^2+(2+2*i)*e^(1/2*τ)*e^(i/2*τ))
Do you prefer the challenge of finding commands that unlock the hidden answer..just curious?

Thanks for replying; there was no insult intended, it seems you took my question as offensive. Using Prime and checking results with other resources in order to provide feedback that can improve the product is a perfectly fine, in fact admirable, reason. But uncommon, hence my curiosity.

And there are no conditions imaginable in which that result could make me happy, correct or not. And, correct or not, any product that provided such an answer surely needs improving. It's a good thing folks like you are helping to prevent folks like me from ever receiving such a result.