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The question is: Evaluate: $$ \lim_{a \to 0}\ \dfrac{ \int^a_0 \ln(1+\tan a\tan x)\ \rm{d}x}{a^3}$$

My method: L'hospital rule.

Let $$L= \lim_{a \to 0}\ \dfrac{ \int^a_0 \ln(1+\tan a \tan x) \ \rm{d}x}{a^3}$$

$$ L= \lim_{a \to 0} \dfrac{ \ln(1+\tan^2a)}{3a^2}$$

$$ \implies L=\dfrac13 $$

The solution given by my teacher :

Let $$L= \lim_{a \to 0}\ \dfrac{ \int^a_0 \ln(1+\tan a \tan x)\ \rm{d}x}{a^3}$$

and let $$ I=\int^a_0 \ln(1+\tan a \tan x)$$

$$ I=\int^a_0 \ln(1+\tan a \tan(a-x)) \ \rm d x$$

$$ I=\int^a_0 \ln\big(1+\tan a \cdot \dfrac{\tan a -\tan x}{1+ \tan a \tan x}\big ) \ \rm d x$$

$$ I= \int^a_0 \ln (1+ \tan^2a) - \ln(1+ \tan a \tan x) \ \rm dx $$

$$ \implies 2I= \int^a_0 \ln (1+ \tan^2a) $$

$$ \implies I= \dfrac{ a \ln ( 1+\tan a^2)}2$$

So, $$ L= \lim_{a \to 0} \dfrac{ \frac a2 \ln( 1+ \tan^2a) \tan^2a}{a^3 \tan^2 a} $$

$$\implies L = \dfrac 12 $$

For me, both methods seem convincing. I don't know where I'm making a mistake.

Help is solicited.Thanks in advance.

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Your method is good and more efficient than your teacher's; but at $\lim_{a \to 0} \dfrac{ \ln(1+\tan^2a)}{3a^2}$ you should apply again l'Hôpital's theorem. – egreg Apr 26 '14 at 14:29
But $ \displaystyle \lim_{a\to 0} \dfrac{ \ln(1+\tan^2 a) } {a^2}= 1 $. – Parth Thakkar Apr 26 '14 at 14:32
Exactly. That's what I'm doing here. – dajoker Apr 26 '14 at 14:33
What is the $\alpha$ doing in the main question? – Sawarnik Apr 26 '14 at 19:07
up vote 6 down vote accepted

You made a mistake the intgral $F(a)=\int_0^a\ln(1+\tan a\tan x)dx$ has $a$ in two places so, the right derivative is $$ F'(a)=\ln(1+\tan^2 a)+\int_0^a\frac{\partial}{\partial a}(\ln(1+\tan a\tan x))dx. $$ That is where your error comes from. You can pursue the calculation as follows $$\eqalign{ F'(a)&=\ln(1+\tan^2 a)+\sec^2a\int_0^a\frac{\tan x}{1+\tan a\tan x}dx\cr &=\ln(1+\tan^2 a)+\sec a\int_0^a\frac{\sin x}{\cos a\cos x+\sin a\sin x}dx\cr &=\ln(1+\tan^2 a)+\sec a\int_0^a\frac{\sin x}{\cos (a-x)}dx\cr &=\ln(1+\tan^2 a)+\sec a\int_0^a\frac{\sin (a-t)}{\cos (t)}dt&(t\leftarrow a-x)\cr &=\ln(1+\tan^2 a)+ \int_0^a(\tan a- \tan t) dt \cr &=\ln(1+\tan^2 a)+ a\tan a+\ln(\cos a)\cr } $$ Now it is easy to see that $$\lim_{a\to0}\frac{F'(a)}{3a^2}=\frac{1}{3}+\frac{1}{3}-\frac{1}{6}=\frac{1}{2}.$$

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Got my mistake. Thanks a lot for your answer. – dajoker Apr 28 '14 at 14:35

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