I have the following integral
\begin{equation} \int_0^1 t^{-5/4} {_1}F_1(-1/4;1/2;t) dt \end{equation}
I know that it only has a pole at zero since the confluent hypergeometric function is analytic in the entire complex plane but I dont know what contour to use.
This integration does not have a convergent result.
$$ \int t^{-5/4} {_1}F_1(-1/4;1/2;t) dt = \int t^{-5/4} \sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k} \frac{t^k}{\Gamma(k+1)} dt \\=\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k} \frac{\int t^{k-\frac{5}{4}}dt}{\Gamma(k+1)} =\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k (k-\frac{1}{4})} \frac{t^{k-\frac{1}{4}}}{\Gamma(k+1)} $$
Since $k-\frac{1}{4} = \frac{3}{4}+k-1 = \frac{\Gamma(-\frac{1}{4})(\frac{3}{4})_k}{\Gamma(\frac{3}{4})(-\frac{1}{4})_k}$
$$=\sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k}\frac{\Gamma(-\frac{1}{4})(\frac{3}{4})_k}{\Gamma(\frac{3}{4})(-\frac{1}{4})_k} \frac{t^{k-\frac{1}{4}}}{\Gamma(k+1)} =\frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4}) \sqrt[4]{t}} \sum_{k = 0}^{+\infty} \frac{(-\frac{1}{4})_k}{(\frac{1}{2})_k}\frac{(\frac{3}{4})_k}{(-\frac{1}{4})_k} \frac{t^{k-\frac{1}{4}}}{\Gamma(k+1)} = \frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4}) \sqrt[4]{t}} {_2}F_2(-\frac{1}{4},-\frac{1}{4};\frac{1}{2},\frac{3}{4};t)$$
When $t \to 1$, $\frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4}) \sqrt[4]{1}} {_2}F_2(-\frac{1}{4},-\frac{1}{4};\frac{1}{2},\frac{3}{4};1) \approx -4.75045$ by wolframalpha.
When $t = 0$, ${_2}F_2(-\frac{1}{4},-\frac{1}{4};\frac{1}{2},\frac{3}{4};0)= 1$ and $\frac{\Gamma(-\frac{1}{4})}{\Gamma(\frac{3}{4})} =-4$. But on the denominator, the $\sqrt[4]{t} =0 $. So it becomes $\frac{-1}{0}$ form, which means the lower limit of integration doesnot exists.
So $\int_0^1 t^{-5/4} {_1}F_1(-1/4;1/2;t) dt$ does not converge.