# Tricky Convergence question

1. Apr 13, 2004

### AngelofMusic

This was a question on one of our past exams. I don't even know where to start and I'm getting very discouraged.

Given $$\int [\frac{1}{\sqrt{1+2x^2}} - \frac{c}{x+1}] = \ln(\frac{\sqrt{1+2x^2} + \sqrt{2}x)}{(x+1)^c}) + k.$$

Find a value c such that $$\int_{0}^{\infty} [\frac{1}{\sqrt{1+2x^2}} - \frac{c}{x+1}]$$ converges.

I tried taking the limit as x-> infinity of the large ln function, but I'm getting nowhere.

Can someone please give me a hint that'll get me started on this question at least?

2. Apr 13, 2004

### ReyChiquito

as x tends to infinitum, the logarithm is superiorly bounded by

$$\frac{1}{(\sqrt{1+2x^2} - \sqrt{2}x)(x+1)^c}$$

and being $$x \geq 0$$ implies that

$$\frac{1}{(\sqrt{1+2x^2} - \sqrt{2}x)(x+1)^c} \leq \frac{1}{(x+1)^c}$$

and this converges if $$c \geq 0$$

Last edited: Apr 13, 2004
3. Apr 14, 2004

### arildno

The term (sqrt(1+2x^(2))-sqrt(2)x)->0 as x->inf; hence you must be more careful in the choice of c.
We should have, in general for e->0, that sqrt(1+e)-1>=1+1/2e-1/8e^(2) (alternating series with terms decreasing in magnitude)
Setting e=1/(2x^(2)), you should find that the expr. converges for c>=1

4. Apr 14, 2004

### arildno

Sorry, estimate all wrong.
You should begin with the original expression.
sqrt(1+2x^(2))+sqrt(2)x=sqrt(2)x(2+1/(4x^(2))-1/(16x^(4))+-+-)->2sqrt(2)x when x->inf.
Hence, convergence requires abs(ln(x^(1-c)))<inf as x->inf.
Only c=1 achieves this result.

5. Apr 14, 2004

### ReyChiquito

true... i dont know what i was thinking... that was my orignial estimate but i messed up resolving the first term of the integral... the shame...

6. Apr 15, 2004

### AngelofMusic

Wow. Thanks a lot! Just one stupid question - how did you get the alternating series expansion for $$\sqrt{1+2x^2} + \sqrt{2}x$$? Did you just take successive derivatives and use Taylor polynomials?

7. Apr 17, 2004

### ReyChiquito

What do you mean?

using the taylor expansion around zero of

$$(1+x)^n \sim 1+nx+n(n-1)\frac{x^2}{2!}+...$$

its easyly shown that

$$\sqrt{1+2x^2} + \sqrt{2}x=\sqrt{2}x(1+\sqrt{1+\frac{1}{2x^2}})=\sqrt{2}x(2+\frac{1}{4x^2}-\frac{1}{16x^4}+...)$$

as x tends to infinitum this function behave like x... the function $$(1+x)^c$$ behaves like $$x^c$$ (same argument) thats why you need $$c \geq 1$$

Last edited: Apr 17, 2004