Determine $\lim_{a\to 0^+} \int _0 ^{\infty} \dfrac{t^{a-1}}{1+t^4} dt$ For $a>0$ we define
$$G(a)= \int _0 ^{\infty} \dfrac{t^{a-1}}{1+t^4} dt$$
Determine $\lim_{a\to 0^+} G(a)$
I don't know really know how to approach this problem that I saw in my measure theory course, without using complex analysis, so I would appreciate any hint. Thank you so much!
 A: There is no need to use a named theorem like DCT or integration by parts, the issue is clearly at 0 since $1/t^4$ is integrable at infinity, and near 0, say $0<t<c=2^{1/4}$, $\frac{1}{1+t^4} ≥ 1/3$. The integrand is positive, so
$$ \int_0^\infty \frac{t^{a-1} dt}{1+t^4} ≥ \frac{1}{3}\int_0^{c} t^{a-1} dt = \frac{c^a}{3a} \xrightarrow[a\to0+]{} \infty$$
A: HINT:
Integration by parts with $u=\frac{1}{1+t^4}$ and $v=\frac{t^a}{a}$ reveals
$$\int_0^\infty \frac{t^{a-1}}{1+t^4}\,dt=\frac1a \int_0^\infty \frac{4t^{a+3}}{(1+t^4)^2}\,dt\geq \frac1a \int_0^\infty \frac{4t^{3}}{(1+t^4)^2}\,dt=\frac{1}{a}.$$

Alternatively, note that 
$$\int_0^\infty \frac{t^{a-1}}{1+t^4}\,dt =\frac{\Gamma(a)\Gamma(4-a)}{\Gamma(4)}\sim \frac1a$$
A: 
Answer:  $$ \color{blue}{\lim_{a\to 0^+}G(a)= \lim_{a\to 0^+}\frac{\pi}{4\sin\frac{ a\pi}{4}} \overset{h= \frac{a\pi}{4}}{=} \lim_{a\to 0^+}\frac{h}{\sin h}}\cdot \frac{1}{a} = \infty~$$

Taking $x=t^4$  we get $dt  =\frac14x^{-\frac34}dx$ $$G(a)= \int _0 ^{\infty} \dfrac{t^{a-1}}{1+t^4} dt =G(a)= \frac14\int _0 ^{\infty} \dfrac{(x^\frac14)^{a-1}}{1+x} x^{-\frac34}dx= \frac14\int _0 ^{\infty} \dfrac{x^{\color{red}{\frac a4-1}}}{(1+x)^{\color{red}{1}}} dx$$
From this $B(x,y) = \int_0^\infty \frac{t^{x-1}}{(1+t)^{x+y}}dt$
 for $0<a<1$ we obtain, By schwarz reflexio formula 
$$G(a) = \frac{1}{4}B\left(\frac{a}{4},1-\frac{a}{4}\right) = \frac{\Gamma\left(\frac{a}{4}\right)\Gamma\left(1-\frac{a}{4}\right)}{4\Gamma (1)} = \frac{\pi}{4\sin\frac{ a\pi}{4}}$$
That is $$ \color{blue}{G(a)= \frac{\pi}{4\sin\frac{ a\pi}{4}} \overset{h= \frac{a\pi}{4}}{=} \frac{h}{\sin h}}\cdot \frac{1}{a} \to \infty~~as~~\to 0^+$$
