Recent content by aristotle_sind

Thank you for the replies... Well .. my bad .. that certainly wasn't a good way to ask for help and my equation looked weird. I've bookmarked the guidelines thread .. will refer to it before making a new thread in future. Thank you for that... And I did get a root that was 5/(2+i) but...

The equation is: (appears while solving a trigonometric integral using residue theorem) 2Z2+iZ2-6Z+2-i =(2+i)Z2-6Z+(2-i) The roots are: Z1=(2-i) and Z2=(2-i)/5 I can't write the equation in factored form. If I simply write it like this: {z-(2-i)}{5z-(2-i)} It doesn't give the same...

Oh .. That clears everything up .. Thanks a lot mate...

I just changed my previous post .. can you please check that...

so I put ##x=-u## in one of the integrals .. this is what I get... \int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = \int_{0}^{R} \frac{\cos mu}{u^2 + 1}\,du + \int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx but now the two integrals on RHS have different variables .. How do I add them so the resulting...

Hello Can someone please tell me how is: \int_{-R}^{R} \frac{\cos mx}{x^2 + 1}\,dx = 2\int_{0}^R \frac{\cos mx}{x^2 + 1}\,dx where, m and R are positive real numbers This is how I'm trying to solve it... \int_{-R}^R \frac{\cos mx}{x^2 + 1}\,dx = \int_{-R}^0 \frac{\cos mx}{x^2 + 1}\,dx...