Hot Integral-5

Calculus Level 5

Let

$\displaystyle f(x)=\int\limits_0^1 \frac{\sqrt{1-xt^2}}{\sqrt{1-t^2}}dt$

Then, the following integral can be expressed as :

$\displaystyle \int \frac{f(2015x^2)}{x^3}dx = -\frac{a}{bz^c}[\pi \left\{dz^f \;_4F_3(g,h,\frac{i}{k},\frac{j}{k};l,m,m;nz^o) -pz^q \log(r) + sz^t\log(-uz^v) + w\right\}]$ + constant

Evaluate $a+b+c+d+f+g+h+i+j+k+l+m+n+o+p+q+r+s+t+u+v+w = ?$

$\textbf{Details and Assumptions}$

1) $a,b,c,d,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w$ are all positive integers.

2) $_4F_3(.,.,.,.;.,.,.;.)$ is a $\textbf{Hypergeometric function}$ .

3) Everything is in the $\textbf{simplest form}$ and $r$ is square free.

$\blacksquare$ This is a part of Hot Integrals

The answer is 12346259.

1 solution

Mark Hennings
Jan 11, 2016

With the substitution $t = \sin\theta$ , the function $f(x)$ becomes $f(x) \; = \; \int_0^{\frac12\pi} \sqrt{1 - x\sin^2\theta}\,d\theta \; = \; E(x) \;,$ the complete elliptic integral of the second kind. Thus $\int \frac{f(2015x^2)}{x^3} \,dx \; = \; \int \frac{E(2015 x^2)}{x^3}\,dx \; = \; \frac{1}{8x^2}G^{2,2}_{3,3}\left(-2015x^2 {\Huge|}{{\frac12,\frac32,2} \atop {0,1,0}}\right)$ using the Meijer G function. This can be expressed as: $-\frac{\pi}{256x^2}\left[ 12180675x^4 {}_4F_3\big(\{1,1,\tfrac32,\tfrac52\};\{2,3,3\};2015x^2\big) - 128960 x^2 \ln 2 + 32240 x^2 \log(-2015x^2) + 64\right]$ making the answer ${\small 1 + 256 + 2 + 12180675 + 4 + 1 + 1 + 3 + 5 + 2 + 2 + 3 + 2015 + 2 + 128960 + 2 + 2 + 32240 + 2 + 2015 + 2 + 64}$ or $12346259$ .

Incidentally, there is a typo in the question. The variable $x$ has morphed into the variable $z$ ...

Right. :) :)

Aman Rajput - 5 years, 5 months ago

Hot Integral-5

The answer is 12346259.

1 solution

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