I want to find the solution to the following integral $$\int_0^{\infty} \frac{dx}{x^{\frac{\alpha}{2}}+1}$$ where $\alpha$ can be any value greater than 2 such that $\alpha /2 >1$ but can be any real value above 1. I have looked at the book "Integral series and products" but the result provided depends on the particular value of $\alpha /2$ that is whether it is even or odd will dictate the result. I want to know if there is some general result for this integral. Thanks in advance.
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This should be a good start: Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer. – Zain Patel Jul 28 '16 at 08:33
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@ZainPatel There are two problems with your answer link. First it considers only integer values. Second, the method described is somehow difficult for me to understand. therefore I will be very thankful if some simpler answer is provided. Thank you. – Frank Moses Jul 28 '16 at 08:42
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Using one of the definitions of the Beta function:
$${\displaystyle \mathrm {B} (a,b)=\int _{0}^{\infty }{\dfrac {t^{a-1}}{(1+t)^{a+b}}}\,\mathrm {d} t,\qquad \mathrm {Re} (a)>0,\ \mathrm {Re} (b)>0\!}$$
We set $$t=x^{\alpha/2}$$
$$dt=\frac{\alpha}{2} x^{\alpha/2-1}dx$$
$$\mathrm {B} (a,b)=\frac{\alpha}{2}\int _{0}^{\infty }{\dfrac {x^{\alpha a/2-1}}{(1+x^{\alpha/2})^{a+b}}}\,\mathrm {d} x$$
$$\alpha a/2-1=0$$
$$a+b=1$$
$$a=\frac{2}{\alpha}$$
$$b=1-\frac{2}{\alpha}$$
$$\int_0^{\infty} \frac{dx}{x^{\frac{\alpha}{2}}+1}=\frac{2}{\alpha} \mathrm {B} \left(\frac{2}{\alpha}, 1-\frac{2}{\alpha} \right)$$
Using the property of Beta function:
$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
We can write:
$$\mathrm {B} \left(\frac{2}{\alpha}, 1-\frac{2}{\alpha} \right)=\Gamma \left(\frac{2}{\alpha} \right) \Gamma \left( 1-\frac{2}{\alpha} \right)$$
And the reflection formula for the Gamma function will get us the result of @robjohn
Yuriy S
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Excellent help. Thank you so much. I found many answer but this is really simpler to comprehend – Frank Moses Jul 28 '16 at 09:10
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yes I have read this in one other post. Further it will make easier if we see that $\Gamma(x) \Gamma(1-x)=\frac{\pi}{sin(\pi x)}$ – Frank Moses Jul 28 '16 at 09:15
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In this answer, it is shown that for $m>0$ and $-1<n<m-1$, $$ \frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x $$ Let $n=0$ and $m=a/2$, to get $$ \int_0^\infty\frac1{1+x^{a/2}}\,\mathrm{d}x=\frac{2\pi}a\csc\left(\frac{2\pi}a\right) $$
