- #1

cjellison

- 18

- 0

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^{p-1}}{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^{p-1}}{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.