Improper Integral $\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$

Asked Sep 04 '16 at 06:37

Active Sep 04 '16 at 10:32

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This integral is from integral

Find $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$$

I have get $$\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx=\int_0^{\infty}\tan\left(\frac1{\sqrt{x+1}}\right)\frac{\ln(1+\sqrt x)}{x}dx$$ Let$$\dfrac{1}{\sqrt{x+1}}=t$$ that $$I=\int_{0}^{1}\dfrac{\tan{t}\ln{\left(1+\sqrt{\frac{1}{t^2}-1}\right)}}{t-t^3}dt$$ This integral is have closed form ?

edited Sep 04 '16 at 10:32

asked Sep 04 '16 at 06:37

math110

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I have to wonder whether this even exists... won't it oscillate too much as you approach zero? – user361424 Sep 04 '16 at 08:02
3

Why is this interesting ? (It's possible to construct infinite complicate integrals.) Something special like e.g. with http://math.stackexchange.com/questions/520657/a-conjectured-closed-form-of-int-0-infty-fracx-1-sqrt2x-1-ln-left2x?rq=1 ? – user90369 Sep 04 '16 at 08:08
@user90369,the same as fell – math110 Sep 04 '16 at 08:40
what does WolframAlpha mean? – Dr. Sonnhard Graubner Sep 04 '16 at 08:54
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Based on past experience with log integrals, I believe that the integral has a better chance of possessing a closed form if $tan$ were replaced with $tan^{-1}$. – Noam Shalev - nospoon Sep 04 '16 at 09:24

Improper Integral $\int_0^\infty\tan\left(\frac x{\sqrt{x^3+x^2}}\right)\frac{\ln(1+\sqrt x)}xdx$

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