For What Values of \(p\) does \(\int_{1}^{\infty} \frac{1}{x^{2} \sqrt{x^{p}-1}} dx \) Converge?

This was a question posed at Mathematics Stack Exchange. The obvious solutions were posted before I saw the question so this is an alternative solution that is interesting yet inefficient.

Let \(y=1/x\), then \(z=y^{-p}-1\)
\begin{align}
\int\limits_{1}^{\infty} \frac{1}{x^{2} \sqrt{x^{p}-1}} dx &= \int\limits_{0}^{1} \frac{1}{\sqrt{y^{-p}-1}} dy \\
&= \frac{1}{p} \int\limits_{0}^{\infty} \frac{z^{-1/2}}{(z+1)^{1+1/p}} dz \\
&= \frac{1}{p} \mathrm{B}\left(\frac{1}{2},\frac{1}{p} + \frac{1}{2} \right) \\
&= \frac{\sqrt{\pi}\Gamma \left( \frac{1}{p} + \frac{1}{2} \right)}{\Gamma \left( \frac{1}{p} \right)}
\end{align}

The integrand and the limits of integration require that \(p \gt 0\). From the
beta function we have \(\mathrm{Re} (\frac{1}{p} + \frac{1}{2}) \gt 0\), \(p \lt -2\) and \(p \gt 0\). Thus we
must have \(p \gt 0\).

We have used the following integral definition of the beta function
\begin{equation}
\mathrm{B}(x+1,y+1) = \int\limits_{0}^{\infty} \frac{z^{x}}{(1+z)^{x+y+2}} dz
\end{equation}

Certain unified integration formulas associated with generalized k-Bessel function by G. Rahman, K. S. Nisar, S. Mubeen, M. Arshad

Our purpose in this present paper is to investigate generalized integration formulas containing the generalized k-Bessel function \(W^{k}_{v,c}(z)\) to obtain the results in representation of Wright-type function. Also, we establish certain special cases of our main result.

The entire paper is available here.

Some integrals involving generalized k-Struve functions by K.S. Nisar, S.R. Mondal

The close form of some integrals involving recently developed generalized k-Struve functions is obtained. The outcome of these integrations is expressed in terms of generalized Wright functions. Several special cases are deduced which lead to some known results.

The entire paper is available here.

New Blog: Advanced Integration: Math, Integration, Special Functions by Zaid Alyafeai

Zaid Alyafeai has started a blog: Advanced Integration: Math, Integration, Special Functions. He is the author of Advanced Integration Techniques, which I highly recommend, and he is active on Mathematics Stack Exchange.

Prove \(\int_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}\)

How to prove
\begin{equation}
\int\limits_{0}^{1} (1-x^{1/k})^{n} dx = {1\over {k+n\choose n}}
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution.

Let \(y=x^{1/k}\)
\begin{align}
\int\limits_{0}^{1} (1-x^{1/k})^{n} dx &= k\int\limits_{0}^{1} (1-y)^{n} y^{k-1} dy \\
&= k\, \mathrm{B}(k,n+1) \\
&= \frac{k\,\Gamma(k)\Gamma(n+1)}{\Gamma(k+n+1)} \\
&= \frac{k!n!}{(k+n)!} \\
&= {1\over {k+n\choose n}}
\end{align}

Evaluate the Integral \(\int_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}} \)

How to evalute
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution

The integral is of the form of the polynomial under the radical having 1 real and 2 complex roots and the lower limit of integration equal to the real root. So we have
\begin{equation}
\int\limits_{a}^{x} \left( (t-a)[(t-r)^{2} + s^{2}] \right)^{-1/2}\, dt
= \frac{1}{\sqrt{m}} \mathrm{cn}^{-1}\left(\frac{m-(t-a)}{m+(t-a)},k \right)
\end{equation}
where
\begin{equation}
m^{2} = (r – a)^{2} + s^{2} \quad \mathrm{and} \quad k^{2} = \frac{1}{2} – \frac{a-r}{2m}
\end{equation}
\(\mathrm{cn}^{-1}z\) is the inverse of one of the Jacobi elliptic functions and \(k\) is the modulus.

To factor the polynomial inside of the radical, we note that the roots are the cube roots of 1, thus
\begin{align}
t^{3}-1 &= (t-t_{1})(t-t_{2})(t-t_{3}) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]-i\frac{\sqrt{3}}{2} \right) \left(\Big[t+\frac{1}{2}\Big]+i\frac{\sqrt{3}}{2} \right) \\
&= (t-1) \left(\Big[t+\frac{1}{2}\Big]^{2} + \frac{3}{4}\right)
\end{align}

We now have
\begin{equation}
a = 1 \quad r = -\frac{1}{2} \quad s = \frac{\sqrt{3}}{2} \quad m^{2} = 3 \quad k^{2} = \frac{1}{2} – \frac{3}{4\sqrt{3}}
\end{equation}
and thus
\begin{equation}
\int\limits_{1}^{x} \frac{dt}{\sqrt{t^{3}-1}}
= 3^{-1/4} \mathrm{cn}^{-1}\left(\frac{\sqrt{3} + 1 – x}{\sqrt{3} – 1 + x},\sqrt{\frac{1}{2} – \frac{3}{4\sqrt{3}}} \right)
\end{equation}

Evaluate the Integral \( \int x^{2} \mathrm{e}^{-x^{2}} dx \)

How to evaluate
\begin{equation}
\int x^{2} \mathrm{e}^{-x^{2}} dx
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution

Integrate by parts

\begin{equation}
\int x^{2} \mathrm{e}^{-x^{2}} dx = \frac{1}{2} \sqrt{\pi} x^{2} \mathrm{erf}(x) – \sqrt{\pi} \int x \, \mathrm{erf}(x) dx
\end{equation}

Integrate by parts again
\begin{align}
\int x \,\mathrm{erf}(x) dx
&= x^{2} \mathrm{erf}(x) + \frac{1}{\sqrt{\pi}} x \mathrm{e}^{-x^{2}}
– \int x \,\mathrm{erf}(x) dx – \frac{1}{\sqrt{\pi}} \int \mathrm{e}^{-x^{2}} dx \\
&= x^{2} \mathrm{erf}(x) + \frac{1}{\sqrt{\pi}} x \,\mathrm{e}^{-x^{2}}
– \int x \mathrm{erf}(x) dx – \frac{1}{2} \mathrm{erf}(x) \\
&= \frac{1}{2} x^{2} \mathrm{erf}(x) + \frac{1}{2\sqrt{\pi}} x \mathrm{e}^{-x^{2}}
– \frac{1}{4} \mathrm{erf}(x)
\end{align}

Thus we have
\begin{equation}
\int x^{2} \mathrm{e}^{-x^{2}} dx = \frac{\sqrt{\pi}}{4} \mathrm{erf}(x) – \frac{1}{2} x \mathrm{e}^{-x^{2}}
\end{equation}

Extension of a factorization method of nonlinear second order ODEs with variable coefficients by H.C. Rosu, O. Cornejo-Perez, M. Perez-Maldonado, J.A. Belinchon

The factorization of nonlinear second-order differential equations proposed by Rosu and Cornejo-Perez in 2005 is extended to equations containing quadratic and cubic forms in the first derivative. A few illustrative physics examples are provided.

The entire paper is available here.

Evaluate the Integral \( \int_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx \)

How to evaluate
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx
\end{equation}
was a question posed at Mathematics Stack Exchange. Here is my solution.

Let \(y=x^{4}\)
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{-3/4} dy
= \frac{1}{4} \gamma\left(\frac{1}{4},1 \right)
\end{equation}

Using the same substitution, we also have
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} x^{4} dx = \frac{1}{4} \int\limits_{0}^{1} \mathrm{e}^{-y} y^{1/4} dy
= \frac{1}{4} \gamma\left(\frac{5}{4},1 \right)
\end{equation}

Thus we obtain
\begin{equation}
\int\limits_{0}^{1} \mathrm{e}^{-x^{4}} (1-x^{4}) dx
= \frac{1}{4} \Big[ \gamma\left(\frac{1}{4},1 \right) – \gamma\left(\frac{5}{4},1 \right) \Big]
\approx 0.7256
\end{equation}

We have used the lower incomplete gamma function:
\begin{equation}
\gamma(s,z) = \int\limits_{0}^{z} \mathrm{e}^{-x} x^{s-1} dx
\end{equation}