Improper integral
- Thread starter Eden20
- Start date
D
Deleted member 4993
Guest
Please show us what you have tried and exactly where you are stuck.
Please follow the rules of posting in this forum, as enunciated at:
Please share your work/thoughts about this problem
Attachments
blamocur
Elite Member
- Joined
- Oct 30, 2021
- Messages
- 2,627
I'll read this in bed
BigBeachBanana
Senior Member
- Joined
- Nov 19, 2021
- Messages
- 2,181
I would guess that it converges. There is no easy way to show that.
My guess is based on a similar problem.
\(\displaystyle \int_2^{\infty}\ln(x^3 + 1) - \ln(x^3 - 1) \ dx = 0.2503272784470577\)
The difference between the two integrals is that there is a way to solve this integral.
My guess is based on a similar problem.
\(\displaystyle \int_2^{\infty}\ln(x^3 + 1) - \ln(x^3 - 1) \ dx = 0.2503272784470577\)
The difference between the two integrals is that there is a way to solve this integral.
Turn [math]\int_{1}^{\infty}\sqrt{x^3+1}-\sqrt{x^3-1}\,\text{d}x[/math] into:
[math]\int_{1}^{\infty}\sqrt{x^3}\sqrt{1+\frac{1}{x^3}}-\sqrt{x^3}\sqrt{1-\frac{1}{x^3}}\,\text{d}x[/math] and then:
[math]\int_{1}^{\infty}\sqrt{x^3}\left[\sqrt{1+\frac{1}{x^3}}-\sqrt{1-\frac{1}{x^3}}\right]\text{d}x[/math] and eventually:
[math]\int_{1}^{\infty}x^{3/2}\left[\left(1+\frac{1}{x^3}\right)^{1/2}-\left(1-\frac{1}{x^3}\right)^{1/2}\right]\text{d}x[/math]
As [imath]x\to\infty[/imath], the [imath]\left(1+\frac{1}{x^3}\right)^{1/2}[/imath] and the [imath]\left(1-\frac{1}{x^3}\right)^{1/2}[/imath] can both be expanded using the binomial expansion.
The dominant [imath]1'\text{s}[/imath] are subtracted away, and then the next most dominant terms become a [imath]p[/imath]-integral of the form:
[math]\int_{1}^{\infty}\frac{1}{x^p}\text{d}x[/math]The value of [imath]p[/imath] determines if the integral converges or diverges.
[math]\int_{1}^{\infty}\sqrt{x^3}\sqrt{1+\frac{1}{x^3}}-\sqrt{x^3}\sqrt{1-\frac{1}{x^3}}\,\text{d}x[/math] and then:
[math]\int_{1}^{\infty}\sqrt{x^3}\left[\sqrt{1+\frac{1}{x^3}}-\sqrt{1-\frac{1}{x^3}}\right]\text{d}x[/math] and eventually:
[math]\int_{1}^{\infty}x^{3/2}\left[\left(1+\frac{1}{x^3}\right)^{1/2}-\left(1-\frac{1}{x^3}\right)^{1/2}\right]\text{d}x[/math]
As [imath]x\to\infty[/imath], the [imath]\left(1+\frac{1}{x^3}\right)^{1/2}[/imath] and the [imath]\left(1-\frac{1}{x^3}\right)^{1/2}[/imath] can both be expanded using the binomial expansion.
The dominant [imath]1'\text{s}[/imath] are subtracted away, and then the next most dominant terms become a [imath]p[/imath]-integral of the form:
[math]\int_{1}^{\infty}\frac{1}{x^p}\text{d}x[/math]The value of [imath]p[/imath] determines if the integral converges or diverges.
pka
Elite Member
- Joined
- Jan 29, 2005
- Messages
- 11,971
This is anything but an elementary problem. Yes it does converge to [imath]\approx 2.02336 [/imath]
To find that number must use the [imath]\Gamma (x) [/imath] function.
[imath][/imath][imath][/imath]
Last edited:
Here is a suggestion/hint:
Rewrite the integrand by using the conjugate:
\(\displaystyle \dfrac{2}{ \ \sqrt{x^3 + 1} \ + \ \sqrt{x^3 - 1} \ }\)
Then come up with an appropriate related expression of which to compare it.
Last edited: