A definite integral

Calculus (Integrals, Series)
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Tolaso J Kos
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A definite integral

#1

Post by Tolaso J Kos »

Let $a>1$. Evaluate the integral:

$$\int_1^{a^2} \frac{\ln x}{\sqrt{x}(x+a)}\, {\rm d}x$$
Imagination is much more important than knowledge.
galactus
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Re: A definite integral

#2

Post by galactus »

I think I may have this one.

$$\int_{1}^{a^{2}}\frac{\ln(x)}{\sqrt{x}(x+a)}dx$$

Let $x=t^{2}, \;\ 2dt=\frac{1}{\sqrt{x}}dx$

$$4\int_{1}^{a}\frac{\ln(t)}{t^{2}+a}dt$$

Parts, Let $u=\ln(t), \;\ dv=\frac{1}{t^{2}+a}dt$

$$\underbrace{\frac{\ln(a)\tan^{-1}(\sqrt{a})}{\sqrt{a}}}_{\text{1}}-\frac{1}{\sqrt{a}}\int_{1}^{a}\frac{\tan^{-1}(\frac{t}{\sqrt{a}})}{t}dt$$

OK. The famous integral $\int\frac{\tan^{-1}(u)}{u}du$ has an indefinite closed form related to dilog:

It is $$\frac{i}{2}[Li_{2}(-iu)-Li_{2}(iu)]$$.

Thus, with the integration limits 1 to a, we have:

$$\frac{i}{2\sqrt{a}}\left[Li_{2}(-i\sqrt{a})-Li_{2}(i\sqrt{a})-Li_{2}(-i/\sqrt{a})+Li_{2}(i/\sqrt{a})\right]$$

which reduces to $$-\frac{\pi i}{2}\log(a)\tag{2}$$

Not forgetting the 4 in front of the integral at the start, and putting the pieces 1 and 2 together we obtain:

$$\boxed{4\frac{\ln(a)\tan^{-1}(\sqrt{a})}{\sqrt{a}}-\frac{\pi}{\sqrt{a}}\ln(a)}$$
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Re: A definite integral

#3

Post by Tolaso J Kos »

Hi C,

here are some more solutions.

1st solution

$$\begin{aligned}
\int_{1}^{a^2}\frac{\ln x}{\sqrt{x}\left ( x+a \right )}\, {\rm d}x &\overset{u=\sqrt{x}}{=\! =\! =\!} 4\int_{1}^{a}\frac{\ln u}{u^2 +a}\, {\rm d}u \\
&=\left [ 4 \frac{\arctan \frac{u}{\sqrt{a}}}{\sqrt{a}} \ln u \right ]_1^a - \frac{4}{\sqrt{a}}\int_{1}^{a}\frac{\arctan \frac{u}{\sqrt{a}}}{u}\, {\rm d}u \\
&= \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1}^{a}\frac{\arctan \frac{u}{\sqrt{a}}}{u}\, {\rm d}u\\
&\overset{y=u/\sqrt{a}}{=\! =\! =\! =\! =\!} \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\arctan y}{y}\, {\rm d}y \\
& \overset{y=1/t}{=\! =\! =\!} \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\arctan \frac{1}{t}}{t}\, {\rm d}t \\
& = \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\left [ \frac{\pi/2 - \arctan t }{t} \right ]\, {\rm d}t \\
&= \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{\pi}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{{\rm d}t}{t} \\
&= \frac{4 \arctan (\sqrt{a})\ln a}{\sqrt{a}} - \frac{\pi \ln a}{\sqrt{a}} \\
& = -\ln a \left [ \frac{\pi - 4\arctan \sqrt{a}}{\sqrt{a}} \right ]
\end{aligned}$$

2nd solution (by matha)

$$\begin{aligned}
\int_{1}^{a^2}\frac{\ln x}{\sqrt{x}(x+a)}\, {\rm d}x &\overset{u=\sqrt{x}} {=\! =\! =\!=\!}\;4\int_{1}^{a}\frac{\ln u}{u^2+a}\, {\rm d}u \\
&\overset{u=\sqrt{a}t}{=\! =\! =\! =\!} \;4\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\ln \sqrt{a}t}{a(t^2+1)}\cdot \sqrt{a}\, {\rm d}t \\
&= \frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln \sqrt{a}+\ln t}{t^2+1}\, {\rm d}t\\
&=\frac{2\ln a}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{{\rm d}t}{t^2+1}+ \cancelto{0}{\frac{4}{\sqrt{a}}\int_{1/\sqrt{a}}^{\sqrt{a}}\frac{\ln t}{t^2+1}\, {\rm d}t}\\
&=-\ln a \left [ \frac{\pi - 4\arctan \sqrt{a}}{\sqrt{a}} \right ]
\end{aligned}$$

3rd solution

$$\begin{aligned}
\int_{1}^{a^2}\frac{\ln x}{\sqrt{x}\left ( x+a \right )}\, {\rm d}x&\overset{u=\sqrt{x}}{=\! =\! =\! =\!} \int_{1}^{a}\frac{2u \ln u^2}{u(u^2+a)} \, {\rm d}u \\
&=4\int_{1}^{a}\frac{\ln u}{u^2+a}\, {\rm d}u \\
&\overset{u=a/t}{=}2 \ln a \int_{1}^{a}\frac{{\rm d}u}{u^2 +a} \\
&=2\ln a \left [ \frac{\arctan \frac{u}{\sqrt{a}}}{\sqrt{a}} \right ]_1^a \\
&= 2\ln a \left [ \frac{\arctan \sqrt{a}-\arctan \frac{1}{\sqrt{a}}}{\sqrt{a}} \right ] \\
&= -\ln a \left [ \frac{\pi -4 \arctan \sqrt{a}}{\sqrt{a}} \right ]
\end{aligned}$$

In all of the above solutions the known fact:

$$\arctan x + \arctan \frac{1}{x} = \frac{\pi}{2}, \;\; x>0$$

was used.

I have taken this exercise from the greek forum mathematica.gr
Imagination is much more important than knowledge.
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