# Solving used Real Based Methods: $\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt$

In working on integrals for the past couple of months, I've come across different cases of the following integral:

$$$$I\left(x,a,k,n,m\right) = \int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt$$$$

Where $$x,a\in \mathbb{R}^{+}$$.

Here the method that I've taken is rather simple and I was curious as to other 'Real' Based methods could be employed with this integral? I also believe that with the conditions I've set on the parameters that it is convergent. If I'm able to expand those conditions, could you please advise.

Interested in special cases too!

The method I took:

First I wanted to bring the 'a' out the front:

$$$$I(x,a,k,n,m) = \int_0^x \frac{t^k}{\left(a\left[\left(a^{-\frac{1}{n}}t\right)^n + 1\right]\right)^m}\:dt = \frac{1}{a^m} \int_0^x \frac{t^k}{\left(\left(a^{-\frac{1}{n}}t\right)^n + 1\right)^m}\:dt$$$$ Here let $$u = a^{-\frac{1}{n}}t$$ Thus,

$$$$I(x,a,k,n,m) = \frac{1}{a^m} \int_0^{a^{-\frac{1}{n}}x} \frac{\left(a^{\frac{1}{n}}u\right)^k}{\left(u^n + 1\right)^m}a^{\frac{1}{n}}\:du = a^{\frac{k + 1}{n} - m}\int_0^{a^{-\frac{1}{n}}x} \frac{u^k}{\left(u^n + 1\right)^m}\:du = a^{\frac{k + 1}{n} - m}I(a^{-\frac{1}{n}}x,1,k,n,m)$$$$

From here I will use $$I$$ in place of $$I(x,a,k,n,m)$$ for ease of typing. The next step is to make the substitution $$w = u^n$$ to yield:

$$$$I = a^{\frac{k + 1}{n} - m}\int_0^{ax^n} \frac{w^\frac{k}{n}}{\left(w + 1\right)^m}\frac{\:dw}{nw^{\frac{n - 1}{n}}} = \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_0^{ax^n} \frac{w^{\frac{k + 1}{n} - 1}}{\left(w + 1\right)^m}\:dw$$$$

Here make the substitution $$z = \frac{1}{1 + w}$$ to yield:

\begin{align} I &= \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_1^{\frac{1}{1 + ax^n}} z^m \left(\frac{1 - z}{z}\right)^{\frac{k + 1}{n} - 1}\left(-\frac{1}{z^2}\right) \:dz = \frac{1}{n}a^{\frac{k + 1}{n} - m}\int_{\frac{1}{1 + ax^n}}^1 z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz \\ &= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[\int_0^1 z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz - \int_0^{\frac{1}{1 + ax^n}} z^{m - \frac{k + 1}{n} - 1}\left(1 - z\right)^{\frac{k + 1}{n} - 1}\:dz \ \right] \\ &= \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left( \frac{1}{1 + ax^n}; m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right] \end{align}

Where $$B(a,b)$$ is the Beta Function and $$B(x; a,b)$$ is the Incomplete Beta Function.

And so, we arrive at:

$$$$\int_0^x \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} \left[B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right) - B\left(\frac{1}{1 + ax^n}; m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right]$$$$

Here we observe that for convergence:

$$$$m - \frac{k + 1}{n} \gt 0,\quad \frac{k + 1}{n} \gt 0,\quad n \neq 0$$$$

Note: when $$x \rightarrow \infty$$ we have:

$$$$\int_0^\infty \frac{t^k}{\left(t^n + a\right)^m}\:dt = \frac{1}{n}a^{\frac{k + 1}{n} - m} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)$$$$

Update: Today I realised that we can use this result for another integral:

$$$$\int_0^\infty \frac{\ln(t)}{\left(t^n + 1\right)^m}\:dt$$$$

This is achieved through a simple use of Feynman's Trick. Here we consider the case when $$x \rightarrow \infty$$ and $$a = 1$$. We see that

\begin{align} \frac{d}{dk}\left[ \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt \right]&= \frac{d}{dk}\left[\frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\right] \\ \int_0^\infty \frac{t^k \ln(t)}{\left(t^n + 1\right)^m}\:dt &= \frac{1}{n^2}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\left[\psi^{(0)}\left(\frac{k + 1}{n}\right) - \psi^{(0)}\left(m - \frac{k + 1}{n}\right) \right] \end{align}

Thus, $$$$\lim_{k \rightarrow 0} \int_0^\infty \frac{t^k \ln(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k \rightarrow 0}\frac{1}{n^2}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right)\left[\psi^{(0)}\left(\frac{k + 1}{n}\right) - \psi^{(0)}\left(m - \frac{k + 1}{n}\right) \right]$$$$

And finally:

$$$$\int_0^\infty \frac{ \ln(t)}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n^2}B\left(m - \frac{1}{n}, \frac{1}{n} \right)\left[\psi^{(0)}\left(\frac{1}{n}\right) - \psi^{(0)}\left(m - \frac{1}{n}\right) \right]$$$$

Note: In the case where $$m = 1$$ we arrive:

\begin{align} \int_0^\infty \frac{ \ln(t)}{\left(t^n + 1\right)^1}\:dt &= \frac{1}{n^2}B\left(1 - \frac{1}{n}, \frac{1}{n} \right)\left[\psi^{(0)}\left(\frac{1}{n}\right) - \psi^{(0)}\left(1 - \frac{1}{n}\right) \right] \\ &= \frac{1}{n^2} \Gamma\left(\frac{1}{n} \right)\Gamma\left(1 - \frac{1}{n} \right) \cdot -\pi\cot\left(\frac{\pi}{n}\right) \\ &= \frac{1}{n^2} \frac{\pi}{\sin\left(\frac{\pi}{n}\right)}\cdot -\pi\cot\left(\frac{\pi}{n}\right) \end{align}

Thus:

$$$$\int_0^\infty \frac{ \ln(t)}{t^n + 1}\:dt = -\frac{\pi^2}{n^2} \operatorname{cosec}\left(\frac{\pi}{n} \right)\cot\left(\frac{\pi}{n}\right)$$$$

• damn bro that's a crazy integral – clathratus Dec 31 '18 at 6:07
• It's more that I see it so often I wanted to finally get it out in a closed form so I could be done with solving it 10 different ways haha. Please if you have time look in detail. I do believe it's correct, but I need some wise eyes to have a look over. – user150203 Dec 31 '18 at 6:15
• Would this work for $n=2$, $a=\frac{1+b}{1-b}$ for some $|b|<1$, and some integer $m\geq1$, and $k=2K$ for integer $0\leq K\leq m-1$? I need to know for $$F(m;b)=\int_0^\pi \frac{\mathrm dx}{(1+b\cos x)^m}$$ – clathratus Jan 17 at 18:03
• @clathratus - Given that the integrand is discontinuous at $x = \frac{\pi}{2}$ - do you expect there to be an elementary primitive?. Regardless, if you changed the upper bound to $\frac{\pi}{2}$ then the half-tangent substitution should yield an elementary form. – user150203 Jan 18 at 9:36
• Check out this: math.stackexchange.com/q/3077345/583016 It's also on our document. I have made a desmos page and it seems as if it works. – clathratus Jan 18 at 17:27