2
$\begingroup$

When I was considering derivatives and integrals (to try integration by parts) related with a closed form for $\zeta(3)$ (due to J. Jensen, see the Wikipedia), I've asked to me

Question. How it is possible to calculate $$\int_0^1\frac{\cos(\arctan x)}{\sqrt{x}}dx?$$ Can you justify a closed form for $$\int\frac{\cos(\arctan x)}{\sqrt{x}}dx?$$ Many thanks.


(The integral $\int \frac{\cos(\arctan x)}{1+\sqrt{x}}dx$ seems much more complicated).

$\endgroup$
  • $\begingroup$ $\cos(\arctan x) = 1/\sqrt{1+x^2}$. But I think that still leaves an elliptic integral. $\endgroup$ – B. Goddard Sep 15 '16 at 16:41
  • $\begingroup$ You are welcome @B.Goddard, very thanks much for the substitution. The problem is that I don't know facts about elliptic integrals. $\endgroup$ – user243301 Sep 15 '16 at 16:43
  • $\begingroup$ the result should be $$\frac{4 \Gamma \left(\frac{5}{4}\right)^2}{\sqrt{\pi }}$$ $\endgroup$ – Dr. Sonnhard Graubner Sep 15 '16 at 16:53
  • $\begingroup$ Very thank much always for your answers in this site @Dr.SonnhardGraubner $\endgroup$ – user243301 Sep 15 '16 at 16:55
2
$\begingroup$

$$I:=\displaystyle \int_0^1\frac{\cos\arctan x}{\sqrt{x}}\,\mathrm d x$$ $\displaystyle \cos\arctan x = \frac{1}{\sqrt{x^2+1}}$, so: $$I=\int_0^1\frac{1}{\sqrt{x(x^2+1)}}\,\mathrm d x$$ Now substitute $x=\tan\frac\theta 2$: $$I=\int_0^1 \frac{1}{\sqrt{x(x^2+1)}}\,\mathrm d x=\frac{\sqrt 2}{2} \int_0^{\frac\pi 2} \frac{\mathrm d \theta}{\sqrt{\sin\theta}}=\frac{\sqrt{2}}{4}\beta \left(\frac 14,\frac 12\right)=\boxed{\frac{\Gamma ^2 \left(\frac 14 \right)}{4\sqrt\pi}}$$ This is also equal to $K\left(\frac 12\right)$.

$\endgroup$
  • $\begingroup$ Thanks a lot for your answer, now I study it. $\endgroup$ – user243301 Sep 15 '16 at 17:44
  • $\begingroup$ I've forgot check it as the answer, I am sorry. $\endgroup$ – user243301 Nov 12 '16 at 21:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy