Integrating a quadratic funtion raised to m/n

[mdnazmulh]

I couldn't figure out what to do with this type of integration \int (a x^{2}+ bx +c) ^{\frac{m}{n}}dx here m, n are integer numbers. Integration limit -∞ to +∞.
Will Binomial expansion work?
Please give me some clue.

 

[Pere Callahan]

What makes you think that these integrals converge? The function to be integrated does not even converge to zero as x-> +/- infinity...!

 

[mdnazmulh]

Actually I had to solve an integral \int(z^{2} -8z + 36)^\frac{-3}{2} dz limit - infinity to + infinity in a physics problem. dz is the differential length of a conducting wire extending from -infinity to + infinity.
Can you please help me how can I solve above integral?

 

[Pere Callahan]

Well, the easiest way is probably to look it up it an table of integrals

Wikipedia states that

where s is the square root of x^2-a^2...taking m=0, n=1 should help you.maybe that is not the bsst way to go.

I rather suggest the following; use simple substitutions to reduce the integrand to
<br /> (x^2+1)^{-3/2}<br />

then use x=\sinh(z) to obtain
<br /> \int_{\mathbb{R}}{\frac{dz}{\cosh(z)^2}}<br />

Once you have that, we can think about the next steps

 

[mdnazmulh]

you gave me a good idea to look it up in integral table. I found one http://www.integral-table.com/

I found quite similar integral in no. 40 in the table's list ( sorry it's quite difficult to write that integral along with its result here) . But How can I obtain that result?

 

[Pere Callahan]

have you simplified your integral to something which looks similar to
<br /> \int_{\mathbb{R}}{\frac{dz}{\cosh(z)^2}}<br />

?

 

[mdnazmulh]

I'm puzzled at the substitution method u suggested. 1st would I let,
z^2- 8 z + 36 = u
2z -8 =du/dz
but dz = du/ (2z-8)
see there's a 'z' in the denominator on right hand side. So, letting z^2- 8 z + 36 = u won't work. What I can do now?

 

[Pere Callahan]

Well...x^2-8x+36 = (x-4)^2+20

thus your first substitution would be

y = x-4.

then

w = y/sqrt(20)

and finally

sinh(z)= w.

 

[mdnazmulh]

yeah. Thank u so much . Now I think I can manage the rest of the things.