AoPS Academy
, 
,
, 
, why 
or even expressions of those terms combined as if that would make them defined. I have made this post to answer these questions once and for all, and I politely ask everyone to link this post to threads that are talking about this issue.. It is usually regarded that 
, not because this works numerically but because it is convenient to define it this way. You will see the convenience of defining other undefined things later on in this post.? The issue here is that 
isn't even rigorously defined in this expression. What exactly do we mean by ? Unless the example in question is put in context in a formal manner, then we say that is meaningless.? Suppose that 
. Then we would have 
, absurd. A more rigorous treatment of the idea is that 
does not exist in the first place, although you will see why in a calculus course. So the point is that is undefined.
? An article from brilliant has a good explanation. Alternatively, you can just use a geometric series. Notice that
? Usually this is considered to be an indeterminate form, but I would also wager that this is also undefined.
indeterminate? In analysis (the theory behind calculus and beyond), limits involving an algebraic combination of functions in an independent variable may often be evaluated by replacing these functions by their limits. However, if the expression obtained after this substitution does not provide sufficient information to determine the original limit, then the expression is called an indeterminate form. For example, we could say that is an indeterminate form.
, 
or , obtained by applying the algebraic limit theorem (a theorem in analysis, look this up for details) in the process of attempting to determine a limit, which fails to restrict that limit to one specific value or infinity, and thus does not determine the limit being calculated. This is why it is called indeterminate. Some examples of indeterminate forms are![\[0/0, \infty/\infty, \infty-\infty, \infty \times 0\]](http://latex.artofproblemsolving.com/2/6/0/260f97a1cf1ea8722ff97a29b83bac7bfe83e577.png)
etc etc. So what makes something undefined? In the broader scope, something being undefined refers to an expression which is not assigned an interpretation or a value. A function is said to be undefined for points outside its domain. For example, the function 
given by the mapping 
is undefined for 
. On the other hand, 
is undefined because dividing by is not defined in arithmetic by definition. In other words, something is undefined when it is not defined in some mathematical context. is considered both indeterminate and undefined (but in the context of a limit then it is considered in indeterminate form). Additionally, this notion of something being undefined also means that we can define it in some way. To rephrase, this means that technically, we can make something that is undefined to something that is defined as long as we define it. I'll show you what I mean.![\[\overline{\mathbb{R}}=\mathbb{R}\cup \{-\infty,+\infty\}.\]](http://latex.artofproblemsolving.com/9/9/3/993c78b62ec2c84b5d77daa8e9a8cd6f70fcf904.png)
So instead of treating infinity as an idea, we define infinity (positively and negatively, mind you) as actual numbers in the reals. The advantage of doing this is for two reasons. The first is because we can turn this thing into a totally ordered set. Specifically, we can let 
for each 
which means that via this order topology each subset has an infimum and supremum and 
is therefore compact. While this is nice from an analytic standpoint, extending the reals in this way can allow for interesting arithmetic! In it is perfectly OK to say that,![\begin{align*}
a + \infty = \infty + a & = \infty, & a & \neq -\infty \\
a - \infty = -\infty + a & = -\infty, & a & \neq \infty \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \pm\infty, & a & \in (0, +\infty] \\
a \cdot (\pm\infty) = \pm\infty \cdot a & = \mp\infty, & a & \in [-\infty, 0) \\
\frac{a}{\pm\infty} & = 0, & a & \in \mathbb{R} \\
\frac{\pm\infty}{a} & = \pm\infty, & a & \in (0, +\infty) \\
\frac{\pm\infty}{a} & = \mp\infty, & a & \in (-\infty, 0).
\end{align*}](http://latex.artofproblemsolving.com/6/d/8/6d8fc1eca5c791d430a7168cd3b37a02104fa318.png)
So addition, multiplication, and division are all defined nicely. However, notice that we have some indeterminate forms here which are also undefined,![\[\infty-\infty,\frac{\pm\infty}{\pm\infty},\frac{\pm\infty}{0},0\cdot \pm\infty.\]](http://latex.artofproblemsolving.com/a/0/b/a0b67076efcc014c78b2023be1151c84b57c1503.png)
So while we define certain things, we also left others undefined/indeterminate in the process! However, in the context of measure theory it is common to define 
as greenturtle3141 noted below. I encourage to reread what he wrote, it's great stuff! As you may notice, though, dividing by is undefined still! Is there a place where it isn't? Kind of. To do this, we can extend the complex numbers! More formally, we can define this extension as![\[\mathbb{C}^*=\mathbb{C}\cup\{\tilde{\infty}\}\]](http://latex.artofproblemsolving.com/8/4/e/84ed3d6ab237df970333b87beaef5311caa46185.png)
which we call the Riemann Sphere (it actually forms a sphere, pretty cool right?). As a note, 
means complex infinity, since we are in the complex plane now. Here's the catch: division by is allowed here! In fact, we have![\[\frac{z}{0}=\tilde{\infty},\frac{z}{\tilde{\infty}}=0.\]](http://latex.artofproblemsolving.com/1/6/0/160cb939707708ff4e0930724df2928af11d7fd1.png)
where 
and are left undefined. We also have
Furthermore, we actually have some nice properties with multiplication that we didn't have before. In 
it holds that![\[\tilde{\infty}\times \tilde{\infty}=\tilde{\infty}\]](http://latex.artofproblemsolving.com/8/e/0/8e035b4841d20c5ad671c46355cc667ca01e533b.png)
but 
and 
are left as undefined (unless there is an explicit need to change that somehow). One could define the projectively extended reals as we did with , by defining them as![\[{\widehat {\mathbb {R} }}=\mathbb {R} \cup \{\infty \}.\]](http://latex.artofproblemsolving.com/c/0/4/c044ff37849acad3b8b280b9753539225a1276ca.png)
They behave in a similar way to the Riemann Sphere, with division by also being allowed with the same indeterminate forms (in addition to some other ones).
and ![$f:[a,b]\rightarrow\mathbb{R}$](http://latex.artofproblemsolving.com/6/7/c/67c320754a36aaec178e61f04bfcbac41c0bf675.png)
be a strictly monotonous function such that 
. Show that 
.
for which there exists a scrictly monotonous function 
such that 
is a random variable that takes on only nonnegative integer values, with ![$E[X]=1,$](http://latex.artofproblemsolving.com/e/8/2/e826f5b79a2adeacdaccd3884658e94a3b1f8e22.png)
![$E[X^2]=2,$](http://latex.artofproblemsolving.com/5/9/1/5915ee4a36f8615ddc8075a197f3b44a902936c6.png)
and ![$E[X^3]=5.$](http://latex.artofproblemsolving.com/d/1/1/d11c925af9f188f0140a89a0fbd6c83e1dd4fc12.png)
(Here ![$E[Y]$](http://latex.artofproblemsolving.com/a/2/9/a29ea940a5d54ad2a5d668a596f9db0c88fa1123.png)
denotes the expectation of the random variable 
) Determine the smallest possible value of the probability of the event 
and 
. Define 
for 
as 
, 
. Prove that 
exists and find it's value .
. If 
is rounded to the nearest integer and then divided by 
, what is the remainder?
.
or 
, then 
and
.
.
.![$I=8\arcsin(2\sqrt{s}-1)+C=8\arcsin(2\sqrt[4]{t}-1)+C=8\arcsin(2\sqrt[8]{x}-1)+C$](http://latex.artofproblemsolving.com/5/f/b/5fbe34c824a24e83a97a5fb8c9183f395284960b.png)
.
.
, then 
and
.![$I=\left[e^{t}\frac{\sin t+\cos t}{2} \cdot \ln(\sin t+\cos t)\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}} e^{t}(\sin t+\cos t) \cdot \frac{\cos t-\sin t}{\sin t+\cos t}\ dt$](http://latex.artofproblemsolving.com/1/f/c/1fce20b6f4ffa16f433b0d9e8fe8f7b74529c6c9.png)
.![$I=\left[e^{t}\frac{\sin t+\cos t}{2} \cdot \ln(\sin t+\cos t)\right]_{0}^{\frac{\pi}{2}}-\frac{1}{2}\int_{0}^{\frac{\pi}{2}} e^{t}(\cos t-\sin t)\ dt$](http://latex.artofproblemsolving.com/1/b/2/1b2f757303e3c6d31e5a2dcbb4e72f1665dc4eec.png)
.![$I=\left[e^{t}\frac{\sin t+\cos t}{2} \cdot \ln(\sin t+\cos t)-\frac{1}{2}e^{t}\cos t \right]_{0}^{\frac{\pi}{2}}=\frac{1}{2}$](http://latex.artofproblemsolving.com/b/8/1/b81efb23bdafc70f68288db7cb1ae9c0ff181582.png)
.
.
, then 
or 
,
.
.
.
.
.
.
.
.
and 
.
.
.
.
.
.
.
, then 
and
.
.
.
![$$\lim_{n \to \infty} \int_{0}^{1} \left[ \frac{\lfloor n^{2025} x \rfloor - (n-1) \left( \lfloor nx \rfloor + \lfloor n^2 x \rfloor + \cdots + \lfloor n^{2024} x \rfloor \right)}{n} \right] dx$$](http://latex.artofproblemsolving.com/4/3/f/43f6ea582fa426bf361289660886f58706ff03fa.png)
![$$= \lim_{n \to \infty} \int_{0}^{1} \left[ \frac{n^{2025} x - \{n^{2025} x\} - (n-1) \sum_{k=1}^{2024} (n^k x - \{n^k x\})}{n} \right] dx$$](http://latex.artofproblemsolving.com/7/d/b/7db2f794e186806490a28f41117eaa60564e1e30.png)
![$$= \lim_{n \to \infty} \int_{0}^{1} \left[ \frac{n^{2025} x - \{n^{2025} x\} - (n-1) \left( x \cdot \frac{n^{2025} - n}{n - 1} - \sum_{k=1}^{2024} \{n^k x\} \right)}{n} \right] dx$$](http://latex.artofproblemsolving.com/c/d/e/cde6f013777a71e0391669bfce1f07a74546e5af.png)
![$$= \lim_{n \to \infty} \int_{0}^{1} \left[ \frac{n^{2025} x - \{n^{2025} x\} - x(n^{2025} - n) + (n-1) \sum_{k=1}^{2024} \{n^k x\}}{n} \right] dx$$](http://latex.artofproblemsolving.com/4/f/0/4f0e0cab3fedf0a87b34e5fe87e3e9ac72dec428.png)
![$$= \lim_{n \to \infty} \int_{0}^{1} \left[ - \frac{\{n^{2025} x\}}{n} + x + \frac{n-1}{n} \sum_{k=1}^{2024} \{n^k x\} \right] dx$$](http://latex.artofproblemsolving.com/4/e/a/4ea3e73c574c08c7e1ae2eee7561d0a1385b5fb6.png)
![$$= \lim_{n \to \infty} \left[ -\frac{1}{n} \cdot \frac{1}{2} + \frac{1}{2} + \frac{n-1}{n} \cdot 2024 \cdot \frac{1}{2} \right]$$](http://latex.artofproblemsolving.com/c/e/5/ce5bae4769353dcdc09378e8bf4094987d208149.png)
![$$= \lim_{n \to \infty} \left[ -\frac{1}{2n} + \frac{1}{2} + \frac{2024n - 2024}{2n} \right]$$](http://latex.artofproblemsolving.com/6/9/e/69eaae41b5a5886488ddd0091c7fc83fe8e048a7.png)
![$$= \lim_{n \to \infty} \left[ \frac{1}{2} + 1012 - \frac{1 + 2024}{2n} \right]$$](http://latex.artofproblemsolving.com/7/3/4/734ed976a59cbc17417faa8e2c2e8e47f6737bcb.png)
![$I=\int_{0}^{1} \frac{1}{(x+2)\sqrt[3]{x^{2}-x^{3}}} \ dx=\int_{0}^{1} \frac{1}{(x+2)\sqrt[3]{x^{2}(1-x)}} \ dx$](http://latex.artofproblemsolving.com/7/0/2/70290054e52f6a52a2d0c43ba256ad74b74a52fa.png)
.
![$J=\int \frac{1}{(x+2)\sqrt[3]{x^{2}(1-x)}} \ dx$](http://latex.artofproblemsolving.com/6/2/b/62b06cf3dafc4fd23d9e0d50eee326ab59ed1bca.png)
.
or 
, then![$J=\int \frac{-1}{(t+1)\sqrt[3]{t}} \ dt$](http://latex.artofproblemsolving.com/a/a/b/aaba7f8cc73b3e576d67d6d8681ee1ecf54a8233.png)
.
, then![$J=\int \frac{-3z}{z^{3}+2}\ dz=\int \frac{-3z}{(z+\sqrt[3]{2})(z^{2}-\sqrt[3]{2}z+\sqrt[3]{4})}\ dz$](http://latex.artofproblemsolving.com/7/b/7/7b700ba4e065694913b5d118882027ef2f77727c.png)
,![$J=\int \left[-\frac{\sqrt[3]{4}z+2}{2(z^{2}-\sqrt[3]{2}z+\sqrt[3]{4})}+\frac{\sqrt[3]{4}}{2(z+\sqrt[3]{2})}\right]\ dz$](http://latex.artofproblemsolving.com/f/b/a/fba640677c18f5b86a63eb97338c234f79a8c599.png)
.![$J=-\frac{\sqrt[3]{4}}{2}\int \frac{z+\sqrt[3]{2}}{z^{2}-\sqrt[3]{2}z+\sqrt[3]{4}}\ dx+\frac{\sqrt[3]{4}}{2}\ln(z+\sqrt[3]{2})$](http://latex.artofproblemsolving.com/f/7/0/f70633d34e1b492010c14d19272c4bc5eddb4555.png)
.![$J=-\frac{\sqrt[3]{4}}{2}\int \frac{\frac{1}{2}(2z-\sqrt[3]{2})+\frac{3\sqrt[3]{2}}{2}}{z^{2}-\sqrt[3]{2}z+\sqrt[3]{4}}\ dx+\frac{\sqrt[3]{4}}{2}\ln(z+\sqrt[3]{2})$](http://latex.artofproblemsolving.com/7/0/4/704b2585b494f8269e4b578e5e1b5fa6dfe988e5.png)
.![$J=-\frac{\sqrt[3]{4}}{4}\ln(z^{2}-\sqrt[3]{2}z+\sqrt[3]{4})-\frac{3}{2}\int \frac{1}{z^{2}-\sqrt[3]{2}z+\sqrt[3]{4}}\ dx+\frac{\sqrt[3]{4}}{2}\ln(z+\sqrt[3]{2})$](http://latex.artofproblemsolving.com/2/2/9/229667615ad495ebe699ad6eb1cf236d35913b07.png)
.![$J=-\frac{\sqrt[3]{4}}{4}\ln(z^{2}-\sqrt[3]{2}z+\sqrt[3]{4})-\frac{\sqrt{3}}{\sqrt[3]{2}}\arctan(\frac{\sqrt[3]{4}z-1}{\sqrt{3}})+\frac{\sqrt[3]{4}}{2}\ln(z+\sqrt[3]{2})$](http://latex.artofproblemsolving.com/3/2/4/3247c05be143024150746681ac760cb27dcad48b.png)
.
![$I=\int_{0}^{1} \frac{1}{(x+2)\sqrt[3]{x^{2}-x^{3}}} \ dx=\frac{\sqrt{3}\cdot \sqrt[3]{4} \cdot \pi}{3}$](http://latex.artofproblemsolving.com/3/3/c/33cda29a92af3c7bd84647767174fb8f4b5045e9.png)
.
.![$\left[e^{2\sqrt{x}\tan \sqrt{x}}\right]'=e^{2\sqrt{x}\tan \sqrt{x}}\left[\frac{1}{\sqrt{x}}\tan \sqrt{x}+\frac{1}{\cos^{2}\sqrt{x}}\right]$](http://latex.artofproblemsolving.com/a/d/f/adf700789a4859520064b295d4951ba4bd7ed5cc.png)
,![$\cos^{2}\sqrt{x} \cdot \left[e^{2\sqrt{x}\tan \sqrt{x}}\right]'=\frac{1}{\sqrt{x}}\sin \sqrt{x}\cos \sqrt{x} \cdot e^{2\sqrt{x}\tan \sqrt{x}}+e^{2\sqrt{x}\tan \sqrt{x}}$](http://latex.artofproblemsolving.com/5/b/3/5b34d5741b4c91d108da02ce2a2a60dfb0945669.png)
,![$\cos^{2}\sqrt{x} \cdot \left[e^{2\sqrt{x}\tan \sqrt{x}}\right]'-\frac{1}{\sqrt{x}}\sin \sqrt{x}\cos \sqrt{x} \cdot e^{2\sqrt{x}\tan \sqrt{x}}=e^{2\sqrt{x}\tan \sqrt{x}}$](http://latex.artofproblemsolving.com/0/f/0/0f0a06230274f49fc590c30aedfdcfa7076dff68.png)
,![$\cos^{2}\sqrt{x} \cdot \left[e^{2\sqrt{x}\tan \sqrt{x}}\right]'+\left[\cos^{2}\sqrt{x}\right]' \cdot e^{2\sqrt{x}\tan \sqrt{x}}=e^{2\sqrt{x}\tan \sqrt{x}}$](http://latex.artofproblemsolving.com/9/7/9/97955104440f1ec4065c61a969363f9da54dd9e3.png)
.
.
. We use product to sum to get this into a nicer form, and then note that the two functions produced by that product to sum are symmetric. Therefore, we are left to evaluate 
. We convert this into the real part of 
.
where the sqrt is def w/ branch cut s.t. 
and 
for large 
.
over a closed contour 
consisting of the real axis from 
to 
and the large semicircular arc 
in the upper half-plane. For large , we have that the exponential factor becomes 
which vanishes as 
.
simple poles at 
and 
. We compute the residue at 
and choose the positive root. This residue turns out to be 
. The residue theorem gives 
and taking the real part gives 
.
![\[
\int_{-\infty}^{\infty} \frac{\cos x \cdot \cos\left(\sqrt{x^2 + 2}\right)}{x^2 + 1} \, dx
\]](http://latex.artofproblemsolving.com/c/6/4/c645efe4d52abcaf8912628b0021e85fdad5d8b4.png)
Consider the function:![\[
f(z) = \frac{\cos z \cdot \cos\left(\sqrt{z^2 + 2}\right)}{z^2 + 1}.
\]](http://latex.artofproblemsolving.com/4/3/d/43dc39c40b9f1e451b108567806e90e162166d41.png)
over a semicircular contour in the upper half-plane, consisting of:
to 
, and
of radius in the upper half-plane.![\[
\oint_{\text{Contour}} f(z) \, dz = 2\pi i \cdot \text{Res}(f, i),
\]](http://latex.artofproblemsolving.com/3/9/2/392cf4a52288a04d2af2198aa5b6025fa791b77e.png)
is the only pole of in the upper half-plane. is:![\[
\text{Res}(f, i) = \lim_{z \to i} (z - i) f(z).
\]](http://latex.artofproblemsolving.com/8/8/9/8899424359c3de76e712e84873888ef218e62e43.png)
into :![\[
f(i) = \frac{\cos i \cdot \cos\left(\sqrt{i^2 + 2}\right)}{i^2 + 1}.
\]](http://latex.artofproblemsolving.com/3/a/6/3a66e428709ea53a1e62b05731b08308be87a542.png)
![\[
\text{Res}(f, i) = \frac{\cosh 1 \cdot \cos 1}{2i}.
\]](http://latex.artofproblemsolving.com/3/3/3/333047ac07e34d53cebb01c997c8b8941a2b9de0.png)
![\[
\oint_{\text{Contour}} f(z) \, dz = 2\pi i \cdot \text{Res}(f, i) = 2\pi i \cdot \frac{\cosh 1 \cdot \cos 1}{2i} = \pi \cosh 1 \cos 1.
\]](http://latex.artofproblemsolving.com/e/9/a/e9ab4f354bb9d82d470a162134c6702e6fa7b0ea.png)
, the integral over the semicircular arc vanishes because:![\[
|f(z)| \leq \frac{M}{R^2}
\]](http://latex.artofproblemsolving.com/3/2/b/32b4dce8cca9d1fd7d280483b0188868e4f1a635.png)
, and the length of is 
. Thus:![\[
\left| \int_{C_R} f(z) \, dz \right| \leq \frac{M\pi}{R} \to 0 \quad \text{as} \quad R \to \infty.
\]](http://latex.artofproblemsolving.com/6/1/0/6106a3da46261c6f05aa2c7a0040e6e6039d5e0e.png)
![\[
\int_{-\infty}^{\infty} f(x) \, dx = \pi \cosh 1 \cos 1.
\]](http://latex.artofproblemsolving.com/7/d/2/7d28629e8f279d00b4cee179bf35f34d029d2ec9.png)
will have branch cuts in the complex plane. Also, 
isn't bounded along the semi circular arc, it can grow arbitrarily large.
.
,
or 
,
.
,
is an odd function.
in the first integral and renaming gives
;
.
gives
and 
(or you can do both at once but I did it this way) to get
whichever form you prefer.![$\int_0^1 \left( \frac{\tan^2 x}{\tan x -x} - \frac{3}{x} \right) dx=\left[ \ln (\tan x -x) - 3\ln (x) \right]_0^1 = \ln (3 \tan(1) -3)$](http://latex.artofproblemsolving.com/6/a/b/6ab1232f5aff91f0a52a5195e9a6671b64458b57.png)
,
We then make the substitution 
to get
.
.
.
By Wallis' Trick:
![$\int_{-1}^1 \frac{dx}{e^{\sqrt[3]{x}}-1}=\left( u=\sqrt[3]{x} \right)=3\int_{-1}^0 \frac{u^2 }{e^u-1} du +3\int_0^1 \frac{u^2 }{e^u-1} du=3\int_0^1 \frac{t^2 }{e^{-t}-1} dt +3\int_0^1 \frac{u^2 }{e^u-1} du=$](http://latex.artofproblemsolving.com/2/5/2/252f9664c2f59e51d9abae0fb9ca412da2ca111b.png)
.
.
.
.
.
,
.![$2I=\int_{0}^{1} \left[\arcsin \sqrt{\frac{\sqrt{x}}{\sqrt{x}+\sqrt{1-x}}}+\arcsin \sqrt{\frac{\sqrt{1-x}}{\sqrt{x}+\sqrt{1-x}}}\right]\ dx$](http://latex.artofproblemsolving.com/7/a/9/7a95e6ca4cbd05ce57b7f8be0285ef582cf0426e.png)
.
.![$\int_0^1 \arctan(2^x-1)dx=\frac{1}{\ln 2} \int_0^1 \frac{\arctan u}{1+u}du=\frac{1}{\ln 2} \left( \left[ \arctan u \ln (1+u) \right]_0^1 - \int_0^1 \frac{\ln(1+u)}{1+u^2}du \right) =$](http://latex.artofproblemsolving.com/e/8/9/e896079e25182113107d2daeb778c51756a002f9.png)
![$$= \sum_{k=1}^{\infty} \left[x - \frac{kx^2}{2}\right]_{\frac{1}{k+1}}^{\frac{1}{k}}$$](http://latex.artofproblemsolving.com/2/b/8/2b8f7fe5f608d0c1d3ae98e06402cca7b96d7199.png)
.![$\left[x\cos x+(x-1)\sin x\right]'\ =\ (1-x)\sin x+x\cos x$](http://latex.artofproblemsolving.com/9/d/7/9d726956332952378365ed342021a19b169810c4.png)
,![$\left[x\cos x+(x-1)\sin x\right]\ +\ (1-x)\sin x+x\cos x=2x\cos x$](http://latex.artofproblemsolving.com/1/f/8/1f8a29b4c2f5c6ebf1e261079cde5738769438d0.png)
.![$I=\frac{1}{2}\left[x+\ln(x\cos x+(x-1)\sin x)\ \right] + C$](http://latex.artofproblemsolving.com/f/f/d/ffdeae4d1f4d970e7acc3e0b8474b4d2a42b26f2.png)
.
.
,
and 
,
.
.![$I=\sqrt{2}\int_{1}^{2} \frac{1}{\sqrt{t^{2}-1}}\ dt=\sqrt{2}\left[\ln(t+\sqrt{t^{2}-1}\right]_{1}^{2}=\sqrt{2}\ln(2+\sqrt{3})$](http://latex.artofproblemsolving.com/6/2/0/6208326ce8ba86b2f11717ae3564fed5a5eb3703.png)
.
.
.![$J=2^{99}\left[-\frac{1}{99(e^{2x}+1)^{99}} \cdot e^{98x}+\int \frac{1}{99(e^{2x}+1)^{99}} \cdot 98e^{98x}\ dx\right]$](http://latex.artofproblemsolving.com/6/a/b/6abc154ae66add2b498b9dafb939a1be21143e20.png)
.![$J=\frac{2^{99}}{99}\left[-\frac{e^{98x}}{(e^{2x}+1)^{99}} +98\int \frac{e^{98x}}{(e^{2x}+1)^{99}}\ dx\right]$](http://latex.artofproblemsolving.com/4/c/8/4c8c3e623e0e903dbd1a46e74524f819331344b3.png)
.
![$=\left[ \arctan u (\arctan(\pi u))^2 \right]_0^\infty -2\pi \int_0^{\infty} \frac{\arctan u \arctan (\pi u)}{1+\pi^2u^2}du+2\int_0^{\infty} \frac{\arctan u \arctan \frac{u}{\pi}}{1+u^2}du=$](http://latex.artofproblemsolving.com/8/2/8/828462eee7a16fa0547fcc24a6bda049bdcc6b38.png)
![$\int_{0}^{\infty} \cot^{-1}(px) \cdot \cot^{-1}(qx)\ dx=\frac{\pi}{2}\left[\frac{1}{p}\ln(1+\frac{p}{q})+\frac{1}{q}\ln(1+\frac{q}{p})\right]$](http://latex.artofproblemsolving.com/d/c/8/dc8ef3e1c1d4625cd2b266dffc093875d33f1644.png)
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