MHB Integration of tan(x^2), is it possible?

[Saitama]

This is not a homework problem but something I saw on an another forum. Though the actual problem doesn't require the integral but I am interested to know if it possible to evaluate it.

The indefinite integral cannot be found in terms of elementary functions but is it possible to evaluate:
$$\int_{0}^{\sqrt{\pi}/2} \tan(x^2)dx$$
? (or with different limits?)
I couldn't find anything over the internet and I don't have any idea about the definite integral.

Any help is appreciated. Thanks!

 

[chisigma]

This is not a homework problem but something I saw on an another forum. Though the actual problem doesn't require the integral but I am interested to know if it possible to evaluate it.

The indefinite integral cannot be found in terms of elementary functions but is it possible to evaluate:
$$\int_{0}^{\sqrt{\pi}/2} \tan(x^2)dx$$
? (or with different limits?)
I couldn't find anything over the internet and I don't have any idea about the definite integral.

Any help is appreciated. Thanks!

The McLaurin expasion of the function is...

$\displaystyle \tan x^{2} = x^{2} + \frac{1}{3}\ x^{6} + \frac{2}{15}\ x^{10} + ... + \frac{B_{2 n}\ (-4)^{n}\ (1-4^{n})}{(2 n)!}\ x^{2 (2n+1)} + ...\ (1)$

... where the $B_{2 n}$ are the so called 'Bernoulli Numbers'. From (1) You derive...

$\displaystyle \int_{0}^{x} \tan \theta^{2}\ d \theta = \frac{1}{3}\ x^{3} + \frac{1}{21}\ x^{7} + \frac{2} {165}\ x^{11} + ... + \frac{B_{2 n}\ (-4)^{n}\ (1-4^{n})}{\{2 (2 n+1)+ 1 \}\ (2 n)!}\ x^{2 (2n + 1) +1} + ...\ (2)$

For $x= \frac{\sqrt{\pi}}{2}$ the (2) should converge fast enough...

Kind regards

$\chi$ $\sigma$

 

[Saitama]

The McLaurin expasion of the function is...

$\displaystyle \tan x^{2} = x^{2} + \frac{1}{3}\ x^{6} + \frac{2}{15}\ x^{10} + ... + \frac{B_{2 n}\ (-4)^{n}\ (1-4^{n})}{(2 n)!}\ x^{2 (2n+1)} + ...\ (1)$

... where the $B_{2 n}$ are the so called 'Bernoulli Numbers'. From (1) You derive...

$\displaystyle \int_{0}^{x} \tan \theta^{2}\ d \theta = \frac{1}{3}\ x^{3} + \frac{1}{21}\ x^{7} + \frac{2} {165}\ x^{11} + ... + \frac{B_{2 n}\ (-4)^{n}\ (1-4^{n})}{\{2 (2 n+1)+ 1 \}\ (2 n)!}\ x^{2 (2n + 1) +1} + ...\ (2)$

For $x= \frac{\sqrt{\pi}}{2}$ the (2) should converge fast enough...

Kind regards

$\chi$ $\sigma$

Ah, that is quite above my current level so I have no idea what are those Bernoulli Numbers but anyways, thanks chisigma! :)

I thought there could be a simpler solution to this but it doesn't look so. :P