# Thread: when is this integral convergent

1. ## when is this integral convergent

2. Hint : use the limit comparison test with $\displaystyle (1+t^2)^{-1}$.

To prove the identity is a simple residue calculation using the keyhole contour.

3. $\displaystyle \int^{\infty}_{0} \frac{t^{z-1}}{(t+1)^{2}} \ dt = \int_{0}^{\infty} \int_{0}^{\infty} t^{z-1} e^{-tx} xe^{-x} \ dx \ dt$ (Laplace Transform of $\displaystyle xe^{-x}$)

switch the order of integration

$\displaystyle = \int_{0}^{\infty} x e^{-x} \int_{0}^{\infty} t^{z-1} e^{-tx} \ dt \ dx = \int^{\infty}_{0} xe^{-x} x^{-z} \Gamma(z) \ dx$

$\displaystyle = \Gamma(z) \int^{\infty}_{0} x^{1-z} e^{-x} \ dx = \Gamma(z) \Gamma(2-z)= \Gamma(z) (1-z)\Gamma(1-z) = \frac{\pi (1-z)}{\sin(\pi z)}$ (reflection formula)

4. nevermind

5. Originally Posted by Bruno J.
Hint : use the limit comparison test with $\displaystyle (1+t^2)^{-1}$.
Could you explain in more detail?

6. Originally Posted by Random Variable
Could you explain in more detail?
Let $\displaystyle u(t)=\frac{1}{(1+t)^2}, v(t)=\frac{t^{z-1}}{(1+t)^2}$. We know $\displaystyle \int_0^\infty u(t) dt$ converges absolutely, hence $\displaystyle \int_0^\infty |v(t)|dt$ converges if $\displaystyle \lim_{t\to \infty} |v(t)|/u(t) = \lim_{t\to \infty} |t^{z-1}|$ is finite, which is when...

7. Originally Posted by Bruno J.
Let $\displaystyle u(t)=\frac{1}{(1+t)^2}, v(t)=\frac{t^{z-1}}{(1+t)^2}$. We know $\displaystyle \int_0^\infty u(t) dt$ converges absolutely, hence $\displaystyle \int_0^\infty |v(t)|dt$ converges if $\displaystyle \lim_{t\to \infty} |v(t)|/u(t) = \lim_{t\to \infty} |t^{z-1}|$ is finite, which is when...
Unless I'm totally mistaken, the original integral is convergent for $\displaystyle 0 < z <2$. I'm not sure how that proof shows that.

8. My mistake! What I wrote only proves convergence when $\displaystyle 0<\mbox{Re }z<1$, and indeed the integral is convergent when $\displaystyle 0< \mbox{Re }z < 2$.