Integral Manipulation: Solving Complex Integrals with q>p>-1 and w=\cosh(x)

[Ted123]

If $q>p>-1$ and $w=\cosh(x)$ then how do I get smoothly from: $$\displaystyle \int^{\infty}_1 \sinh^{p-1}(x) w^{-q} \;dw$$ to $$\displaystyle \int^{\infty}_1 (w^2-1)^{\frac{p-1}{2}} w^{-q}\;dw$$ and if $t=w^{-2}$ how do I get smoothly from this to: $$\displaystyle \frac{1}{2} \int^{\infty}_0 t^{\frac{p+1}{2}-1} (1-t)^{\frac{q-p}{2}-1}\;dt$$

 

[tiny-tim]

(you know cosh² = sinh² + 1 ?)

looks straightforward to me

where exactly are you having difficulty?

 

[Ted123]

(you know cosh² = sinh² + 1 ?)

looks straightforward to me

where exactly are you having difficulty?

Ah, that identity gives me the first one.

Now if $t=w^{-2}$ then $\displaystyle dw = -\frac{dt}{2w^{-3}}$ so: $$\displaystyle \int^{\infty}_1 (w^2-1)^{\frac{p-1}{2}} w^{-q}\;dw = -\frac{1}{2} \int^{\infty}_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{q}{2}}\;\frac{dt}{w^{-3}}$$ and since $w^{-3} = t^{\frac{3}{2}}$ this equals $$\displaystyle -\frac{1}{2} \int^{\infty}_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt$$

How do I get from this to $$\displaystyle \frac{1}{2} \int^{\infty}_0 t^{\frac{p+1}{2}-1} (1-t)^{\frac{q-p}{2}-1}\;dt$$

 

[tiny-tim]

Hi Ted123!

(just got up :zzz: …)

You're right , it doesn't work …

I think the question has a misprint, it should be …
$$\displaystyle \int^{\infty}_1 \sinh^{q-p}(x) w^{-q} \;dw$$

 

[Ted123]

Hi Ted123!

(just got up :zzz: …)

You're right , it doesn't work …

I think the question has a misprint, it should be …
$$\displaystyle \int^{\infty}_1 \sinh^{q-p}(x) w^{-q} \;dw$$

I think I've got it to work. When we let $t=w^{-2}$ the limits of the integral change from $\int^{\infty}_1$ to $\int^0_1 = -\int^1_0$.

So from $$\displaystyle \frac{1}{2} \int^1_0 (t^{-1} -1)^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt$$

we have that $t^{-1} - 1 = \frac{1-t}{t}$. Hence we have $$\displaystyle \frac{1}{2} \int^1_0 \left( \frac{1-t}{t} \right) ^{\frac{p-1}{2}} t^{\frac{1}{2}(q-3)}\;dt$$

$$= \frac{1}{2} \int^1_0 (1-t)^{\frac{p-1}{2}} \frac{ t^{\frac{1}{2} (q-3)} }{t^{ \frac{p-1}{2} }} \;dt$$

$$= \frac{1}{2} \int^1_0 (1-t)^{\frac{p-1}{2}} t^{\frac{q-p}{2}-1} \;dt = \frac{1}{2} \int^1_0 (1-t)^{\frac{p+1}{2} - 1} t^{\frac{q-p}{2}-1} \;dt$$ This is the beta function $\displaystyle \frac{1}{2} B \left (\frac{q-p}{2} , \frac{p+1}{2} \right)$ and the Beta function is symmetric so this equals $\displaystyle \frac{1}{2} B \left (\frac{p+1}{2} , \frac{q-p}{2} \right)$ which gives us $$\frac{1}{2} \int^1_0 (1-t)^{\frac{q-p}{2}-1} t^{\frac{p+1}{2} - 1} \;dt$$