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Problem :

Find the value of $\int^{\pi/2}_0 \frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$

Now how to use Leibnitz's rule : $\frac{d}{dx}(\int^{\psi(x)}_{\phi(x)} f(t)dt) = \frac{d}{dx}\{\psi(x)\} f(\psi(x)) -\frac{d}{dx}\{\phi(x)\} f(\phi (x)\}$ here

please guide will be of great help thanks as there is $x$ and $\theta$ both

Apurv
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Sachin
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  • Differentiate $\displaystyle f(x) = \int^{\pi/2}_0 \frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$ wrt $x$, then $\displaystyle f'(x) = \int^{\pi/2}_0 \frac{1}{(1+x\sin^2\theta)}d\theta$, and then put $t = \cot \theta$ – r9m Jun 08 '15 at 16:44
  • @r9m Shouldn't it be $f'(x)=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+x\sin^{2}(\theta)}d\theta$? – user99163 Jun 08 '15 at 16:46
  • @r9m ; That's not quite right. Differentiation w.r.t $x$ should yield a $\sin ^2\theta$ in the numerator that will cancel with the one in the denominator. – FH93 Jun 08 '15 at 16:46
  • @user99163 thanks .. that was a mistake! – r9m Jun 08 '15 at 16:47
  • @r9m You're welcome. – user99163 Jun 08 '15 at 16:48

4 Answers4

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Let: $$ f(x) = \int_{0}^{\pi/2}\frac{\log(1+x\sin^2\theta)}{\sin^2\theta}\,d\theta.\tag{1}$$ We have $f(0)=0$ and for every $x\geq 0$: $$ f'(x) = \int_{0}^{\pi/2}\frac{d\theta}{1+x\sin^2\theta}=\frac{\pi}{2\sqrt{1+x}}\tag{2}$$ through Weierstrass substitution. Integrating back, $$ f(x) = \pi\left(\sqrt{1+x}-1\right).\tag{3}$$

Jack D'Aurizio
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In the interests of alternative methods: $$\int_0^{\pi/2}\frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta =-\sum_{k=1}^\infty\frac{(-1)^kx^k}{k}\int_0^{\pi/2}\sin^{2k-2}\theta d\theta.$$ Now let $u=\sin\theta$, then $\frac{du}{\sqrt{1-u^2}}=d\theta$ and we have $$-\sum_{k=1}^\infty\frac{(-1)^kx^k}{k}\int_0^1\frac{u^{2k-2}}{\sqrt{1-u^2}}du = -\frac{\sqrt{\pi}}{2}\sum_{k=1}^\infty\frac{(-1)^kx^k}{k}\frac{\Gamma\left(k-\frac{1}{2}\right)}{\Gamma(k)}.$$ Mathematica quickly evaluates this to have closed form $$\pi\left(\sqrt{x+1}-1\right),$$ hence we have $$\int_0^{\pi/2}\frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta=\pi\left(\sqrt{x+1}-1\right).$$

There may be conditions on $x$ and the exchange of integral and summation would need to be justified. Proof of the summation would also be required of course.

pshmath0
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Let $I(x)$ be the integral

$$I(x)=\int_0^{\pi/2}\frac{\log(1+x\sin^2 \theta)}{\sin^2\theta}d\theta$$

and assume that $x>0$ is real-valued.

Now, taking a derivative with respect to $x$ gives

$$I'(x)=\int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta=\frac{\pi}{2\sqrt{1+x}}$$

where this latter integral was evaluated using contour integration.

Next, integration of $I'(x)$ gives

$$I(x)=\pi\sqrt{1+x}+C$$

where $C$ is found by noting $I(0)=0$. Thus, $C=-\pi$.

Finally, we have

$$I(x)=\pi(\sqrt{1+x}-1)$$

NOTE:

Here, we will evaluate $\int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta$ for $x<0$ using contour integration. Note that in general, we have

$$\begin{align} \int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta & = \int_0^{\pi/2}\frac{1}{1+x\left(\frac{1-\cos 2\theta}{2}\right)}d\theta \tag 1 \\\\ &=-\frac{1}{2x}\int_0^{2\pi}\frac{1}{\cos \theta-a}d\theta \tag 2 \end{align}$$

where $a=\frac{2+x}{x}=1+\frac2x$. In going from $(2)$ to $(3)$ we exploited the periodicity of the cosine function and effected the substitution $2x \to x$.

Next, we move to the complex plane by letting $z=e^{i\theta}$ so that $dz=ie^{i\theta}dz$ and the integration is over the closed-contour unit circle$C$, for which $|z|=1$. Then, we have

$$\begin{align} -\frac{1}{2x}\int_0^{2\pi}\frac{1}{\cos \theta-a}d\theta &=-\frac{1}{ix}\oint_C \frac{dz}{z^2-2az+1}\\\\ &=-\frac{2\pi}{x}\text{Res}\left(\frac{1}{(z-a+\sqrt{a^2-1})(z-a-\sqrt{a^2-1})}\right)\\\\ &=-\frac{2\pi}{x}\text{Res}\left(\frac{1}{(z-\frac{x+2}{x}-\frac{2\sqrt{x+1}}{|x|})(z-\frac{x+2}{x}+\frac{2\sqrt{x+1}}{|x|})}\right) \end{align}$$

There are three cases we need to examine.

Case 1: $x>0$

For $x>0$, the only pole is at $z=\frac{x+2-2\sqrt{x+1}}{x}$. The residue is thus $-\frac{x}{4\sqrt{x+1}}$ and we have

$$I'(x)=\frac{\pi}{2\sqrt{x+1}}$$

Case 2:

For $<-1x<0$, the only pole is at $z=\frac{x+2}{x}+2\frac{\sqrt{x+1}}{|x|}=\frac{x+2-2\sqrt{x+1}}{x}$. The residue is $-frac{x}{4\sqrt{x+1}} and we have

$$I'(x)=\frac{\pi}{2\sqrt{x+1}}$$

Case 3:

For $x<-1$, we note that $|\frac{x+2}{x}+2\frac{\sqrt{x+1}}{|x|}|=1$ and the poles are complex conjugates residing on the unit circle. Thus, the integral diverges.

We can intepret, however, a Cauchy Principal value of the integral. Here, we exclude the poles by modifying the contour $C$ with a semi-circle deformation around each pole. The two contributions from integration around the deformations are equal in magnitude and of opposite sign. Thus, we have

$$I'(x)=0$$

Mark Viola
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$$I(x) = \int_0^{\pi/2}\frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$$

$$I'(x) = \int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta$$

Writing 1 as $\sin^2\theta + \cos^2\theta$, we have

$$\int_0^{\pi/2}\frac{d\theta}{\sin^2\theta(1+x) +\cos^2\theta}$$

Multiplying and dividing by $\sec^2\theta$ and setting $\sqrt{1+x}\tan\theta = t$,

We evaluate the integral to be

$$I'(x)=\frac{\pi}{2\sqrt{x+1}}$$

Then by elementary integration, we get the desired result, and checking that $I(0) = 0$

$$I(x) = \pi(\sqrt{1+x}-1)$$

Mystik141
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