 #1
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Hi, I've typed up my work. Please see the attached pdf.
Basically, I am trying to sovle this problem.
[tex]
\int_0^\infty \frac{x^\alpha}{x^2+b^2} \mathrm{d}x
[/tex]
for [itex]0 <\alpha < 1[/itex]. I follow the procedure given in Boas pg 608 (2nd edition)...and everything seems to work. However, when I check my results, with mathematica for various values of [itex]\alpha[/itex] and [itex]b[/itex], I get incorrect results.
Something I noticed: If I take the absolute value of my generally complex answer, I always get the correct (Mathematica) answer.
So, I would like someone to explain why this happens. What mistake have I made in my solution, and why doesn't this mistake effect Boas in her solution to:
[tex]
\int_0^\infty \frac{r^{p1}}{1+r} \mathrm{d}r
[/tex]
Both problems are solved in the same way...yet her solution [itex] \pi / \sin(\pi p)[/itex] is valid and does not require you to take the absolute value of it.
Thanks so much.
Basically, I am trying to sovle this problem.
[tex]
\int_0^\infty \frac{x^\alpha}{x^2+b^2} \mathrm{d}x
[/tex]
for [itex]0 <\alpha < 1[/itex]. I follow the procedure given in Boas pg 608 (2nd edition)...and everything seems to work. However, when I check my results, with mathematica for various values of [itex]\alpha[/itex] and [itex]b[/itex], I get incorrect results.
Something I noticed: If I take the absolute value of my generally complex answer, I always get the correct (Mathematica) answer.
So, I would like someone to explain why this happens. What mistake have I made in my solution, and why doesn't this mistake effect Boas in her solution to:
[tex]
\int_0^\infty \frac{r^{p1}}{1+r} \mathrm{d}r
[/tex]
Both problems are solved in the same way...yet her solution [itex] \pi / \sin(\pi p)[/itex] is valid and does not require you to take the absolute value of it.
Thanks so much.
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